torsion theories and auslander-reiten sequences · 2016-06-17 · ance, and who is currently...

Torsion theories and Auslander-Reiten

sequences

Puiman Ng

Thesis submitted for the degree ofDoctor of Philosophy

School of Mathematics & StatisticsNewcastle UniversityNewcastle upon Tyne

United Kingdom

February 2011

The Road goes ever on and on

Down from the door where it began.

Now far ahead the Road has gone,

And I must follow, if I can,

Pursuing it with eager feet,

Until it joins some larger way

Where many paths and errands meet.

And whither then? I cannot say.

J. R. R. Tolkien

The Lord of the Rings

i

Acknowledgements

The author would like to express her thankfulness for the financial assistancefrom the School of Mathematics and Statistics, Newcastle University, forthe Newcastle University International Postgraduate Scholarship (NUIPS),for the Overseas Research Students Awards Scheme (ORSAS) Award, andfor the Croucher Foundation Scholarship, which have enabled her study inNewcastle University. The author would like to express her thankfulnessto R.B.J.T. Allenby and John Truss, who have both been enriching andnourishing in their touch of British sense of humour, which has been inter-nalized and absorbed into the author’s being and has graced her nature.The author would like to express her thankfulness to Rafael Bocklandt andVanessa Miemietz for being her examiners, who have read the thesis withutmost conscientiousness and profound thoroughness, and who have stimu-lated her understanding during the viva and who have offered a few gentlesuggestions, including an example which illustrates that not all torsion the-ories are t-structures (Example 4.5.8). The author would like to expressher thankfulness to Yann Palu for suggesting the constructions of mappingcones of (compositions of) downward morphisms in Db(mod kA7) and in thecluster category of Dynkin type A∞ (Section A.3.1 and Section A.3.2). Theauthor would like to express her thankfulness to Robert Marsh for readingthe very first draft of the thesis, who is always subtle in his support and guid-ance, and who is currently reading the book Asterix in Switzerland ([12]).The author would like to express her thankfulness to Agustın Fernandez whohas always been and shall always be the source of the author’s inspiration.The author would also like to express her special gratitude to her supervi-sor, Peter Jørgensen, for his immense ability in seeing the sophistication insimple things and his elegance of simplicity in his unique and individual artand mastery of telling a story.

ii

Abstract

Chapter 0 gives a gentle background to the thesis. It begins with some gen-eral notions and concepts from homological algebra. For example, not onlyare the notions of universal property and of duality central to the flavourof the subject, they are also suggestive in understanding mathematics atanother depth. In category theory, objects and morphisms are the two mainelements in a category, and notions such as kernels and cokernels are definedin terms of objects together with morphisms. In accordance with it, the mor-phisms are given a very subtle significance within a category. The chapterthen introduces the notion of a triangulated category, where due to the lackof uniqueness of certain morphisms described in the axioms, is allowed to befar from an abelian category. A few examples of triangulated categories aregiven, the homotopy category, the derived category and certain stable cat-egories. The chapter ends with a little description of an Auslander-Reitenquiver defined on a Krull-Schmidt category, as well as the notions of Serrefunctor and of Auslander-Reiten triangles in subcategories. The introduc-tion chapter selects lemmas and theorems not only to be referenced in laterchapters, but also those which can induce good intuition on the reader, forexample, in their capacity of being analogues to each other, in the inter-play between them and in their different suggestiveness in approximating orgeneralizing concepts in different ways and directions.

Chapter 1 studies torsion pairs in abelian categories and torsion theorieswith torsion theory triangles in triangulated categories. It then gives anecessary and sufficient condition for the existence of certain adjoint functorsin triangulated categories. Intuitively, they are all different expressions ofsubcategories approximating their ambient categories. The chapter goeson to introduce two special cases of torsion theories, namely t-structuresand split torsion theories, and finishes with a characterization of a splittorsion theory and a classification of split torsion theories in a chosen derivedcategory.

There is a very close and subtle relationship between the existence of torsiontheory triangles and the existence of Auslander-Reiten triangles. Chapter2 studies the existence of Auslander-Reiten sequences in subcategories ofmod(Λ), where Λ is a finite-dimensional k-algebra over the field k, based onthe theory of the existence of Auslander-Reiten triangles in subcategoriesdeveloped by Jørgensen. The existence theorems strengthen the results byAuslander and Smalø and by Kleiner.

Chapter 3 sees that quotients of certain triangulated categories are triangu-lated and are in addition derived categories, appealing to a theorem whichis a slight variation of the results by Rickard and Keller. In this chapter,the Auslander-Reiten triangles play a predominant role in reflecting the tri-

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angulation structure of a triangulated category, and the Auslander-Reitentriangles can be read off from the Auslander-Reiten quiver.

The cluster category D of Dynkin type A∞ was introduced by Jørgensen.One of its several definitions, which is completely analogous to the definitionof the cluster category of type An, motivates us to say that D is a clustercategory of type A∞. In the result by Holm and Jørgensen, the clustertilting subcategories of D were shown to be in bijection with certain maximalsets of non-crossing arcs connecting non-neighbouring integers. Chapter 4generalizes the result by giving a bijection between torsion theories in D andcertain configurations of arcs connecting non-neighbouring integers. Finally,a few examples, characterizing all t-structures and co-t-structures in D, aregiven.

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Contents

0 Introduction 1

0.1 Homological algebra . . . . . . . . . . . . . . . . . . . . . . . 1

0.2 Triangulated categories . . . . . . . . . . . . . . . . . . . . . 15

0.2.1 The homotopy category . . . . . . . . . . . . . . . . . 18

0.2.2 The derived category . . . . . . . . . . . . . . . . . . . 21

0.2.3 The stable category . . . . . . . . . . . . . . . . . . . 33

0.3 Auslander-Reiten quiver . . . . . . . . . . . . . . . . . . . . . 34

1 The approximation properties of subcategories 38

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.2 Existence of certain adjoint functors . . . . . . . . . . . . . . 39

1.3 Examples of torsion theories . . . . . . . . . . . . . . . . . . . 46

1.4 The finite derived category of Dynkin type . . . . . . . . . . . 48

1.5 Anecdote . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2 Existence of Auslander-Reiten sequences in subcategories 59

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.2 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2.3 Weakened notions of precovers (preenvelopes) . . . . . . . . . 62

2.4 Existence of Auslander-Reiten sequences in subcategories . . 70

2.5 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3 Quotients of certain triangulated categories as derived cat-

v

egories 77

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

3.2 The finite derived category Db(mod kAn) . . . . . . . . . . . 78

3.2.1 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.2.2 Mapping cone construction . . . . . . . . . . . . . . . 85

3.2.3 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 88

3.3 The cluster category of Dynkin type A∞ . . . . . . . . . . . . 88

3.3.1 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . 91


3.3.3 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 101

3.4 The finite derived category Db(mod kD5) . . . . . . . . . . . 101

3.4.1 Hom spaces . . . . . . . . . . . . . . . . . . . . . . . . 105

3.4.2 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . 108


3.4.4 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 119

4 A characterization of torsion theories in the cluster categoryof Dynkin type A∞ 120

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

4.2 Coordinate system . . . . . . . . . . . . . . . . . . . . . . . . 121

4.3 Precovering (preenveloping) subcategories . . . . . . . . . . . 126

4.4 Torsion theories . . . . . . . . . . . . . . . . . . . . . . . . . . 130

4.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

A 153

A.1 Coherence problems for adjoint functors . . . . . . . . . . . . 153

A.2 Coherence problems for Serre functors . . . . . . . . . . . . . 155

A.3 Mapping cone construction . . . . . . . . . . . . . . . . . . . 161

A.3.1 The finite derived category Db(mod kA7) . . . . . . . 161

A.3.2 The cluster category of Dynkin type A∞ . . . . . . . . 168

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Chapter 0

Introduction

The chapter serves as a prelude to the thesis. The definitions, lemmas andtheorems are not introduced once and for all, but they will be reincarnatedagain and again in the subsequent chapters, so that each time they appearwith a unique signification and acquire a life of their own, depending onthe context where they manifest themselves, thus altogether a unique phe-nomenon each time. The totality of the thesis can only be attained whenall the chapters are put together like an orchestra, serving a greater whole.Let us now begin.

0.1 Homological algebra

In this section, the reader is suggested to refer to [18] for the background.

Definition 0.1.1. A category C consists of

(i) a class of objects X,Y, Z, . . .,

(ii) for any pair of objects X, Y in C, a set HomC(X,Y ), or simply (X,Y ),when the underlying category is understood, of morphisms from X toY ,

(iii) for objects X, Y , Z in C, a composition relation (X,Y ) × (Y,Z) →(X,Z).

The morphism f from X to Y is written f : X → Y . The set (X,Y )×(Y, Z)consists of pairs (f, g) where f : X → Y , g : Y → Z and the composition off and g is written gf . In addition, C is to satisfy the following axioms.

(i) the sets (X1, Y1) and (X2, Y2) are disjoint unless X1 = X2 and Y1 = Y2,

1

(ii) given the morphisms f : W → X, g : X → Y and h : Y → Z, therelation h(gf) = (hg)f ,

(iii) for each object X there is an identity morphism 1X : X → X, some-times written idX , such that for any f : X → Y and g : W → X, wehave f1X = f and 1Xg = g. When the object X is understood, theidentity morphism is simply written 1 or id.

Given a category C, let Cop be the opposite category. The objects of Cop and Care the same with HomCop(X,Y ) = HomC(Y,X). The composition relationin Cop follows naturally from C, i.e. given morphisms fop in HomCop(Z, Y ),gop in HomCop(Y,X), with the corresponding morphisms f in HomC(Y,Z),g in HomC(X,Y ), the composition gopfop in HomCop(Z,X) takes the valuefg in HomC(X,Z).

A morphism f : X → Y in C is an isomorphism if and only if there is amorphism g : Y → X in C such that gf = 1X and fg = 1Y , and we writeg = f−1 and X ∼= Y . A morphism f : X → X in C is an endomorphismand an isomorphism f : X → X in C is an automorphism. A zero object 0in C is one such that, for any object X in C, the sets (X, 0) and (0, X) bothconsist of precisely one element. Any zero objects in C, if they exist, areisomorphic.

Any class of morphisms M together with a composition relation is sufficientto define a category C. Then a subclass of morphisms M0 of M is to definea subcategory C0 of C if for every f, g in M0, if gf is defined in M, then gfis in M0, and given any identity morphism e in M and any morphism f inM0, and if either ef or fe is defined in M, then e is in M0. In particular,a full subcategory C0 of a category C satisfies HomC0(X,Y ) = HomC(X,Y )for any objects X,Y of C0. Therefore to give a full subcategory C0 of C,it suffices to specify its objects. This notion illustrates the importance ofmorphisms in a category. For example, if a morphism is an isomorphism inC, then it is also an isomorphism in the full subcategory C0 of C.

A covariant (resp. contravariant) functor F : C → D assigns an objectFX in D for each object X in C, and a morphism Ff in HomD(FX,FY )(resp. HomD(FY, FX)) for each morphism f in HomC(X,Y ), such thatF (fg) = (Ff)(Fg) (resp. F (fg) = (Fg)(Ff)) and F (1A) = 1FA for eachobject A in C. A (covariant) functor F : C→ D is full if given objects X,Yin C, the map FXY : HomC(X,Y )→ HomD(FX,FY ) is onto, and is faithfulif given objects X,Y in C, the map FXY : HomC(X,Y )→ HomD(FX,FY )is injective. Finally, F is a full embedding if F is full, faithful and injectiveon objects.

Let Λ be a finite-dimensional k-algebra over the field k. A left module overΛ, or a Λ-left-module, is an abelian group A with a scalar multiplication ·

2

such that given λ in Λ and a in A, the element λ · a is in A, and givena, a1, a2 in A, λ, λ1, λ2 in Λ, the following axioms are satisfied.

(i) (λ1 + λ2) · a = λ1 · a+ λ2 · a,

(ii) (λ1λ2) · a = λ1 · (λ2 · a),

(iii) 1Λ · a = a, where 1Λ 6= 0 is the unity element of Λ,

(iv) λ · (a1 + a2) = λ · a1 + λ · a2.

A Λ-left-submodule is a subgroup A′ of A with λ · a′ in A′ for all λ in Λand a′ in A′. A Λ-right-(sub)module is defined similarly. A Λ-bi-module isa module which is simultaneously a Λ-left-module and a Λ-right-module.

Let A and B be Λ-left-modules. A map f : A → B is a homomorphism ifgiven u, v in A and λ in Λ, we have f(u+v) = f(u)+f(v) and f(λu) = λf(u).Let S be a subset of A, and let A0 be the set of all elements a in A of theform a = Σλss where λs is in Λ and λs 6= 0 for only a finite number ofelements s in S. It is trivial that A0 is a submodule of A. If the submoduleA0 is equal to A, then S is a set of generators of A. If A admits a finite setof generators, it is said to be finitely generated. A set S of generators of Ais said to be a basis of A if every element a of A can be expressed uniquelyin the form a = Σλss where λs is in Λ and λs 6= 0 for only a finite numberof elements s in S. If S is a basis of A, then A is said to be free on the setS. If A is free on some subset, then A is said to be free.

Example 0.1.2. The category Mod(Λ) is a category with Λ-left-modules asobjects and homomorphisms as morphisms. The category mod(Λ) is a cate-gory with finitely generated Λ-left-modules as objects and homomorphismsas morphisms.

Below are a few definitions in an additive category. They are deeply involvedin an abelian category. Both notions will be introduced in a little while. Akernel of a morphism ϕ : A→ B, denoted by kerϕ, is a morphism µ : K → Asuch that µϕ = 0 and if ϕψ = 0 for ψ : X → A, then there is a uniqueψ′ : X → K such that ψ = µψ′. This is suggestive of the following dualnotion. A cokernel of a morphism ϕ : A → B, denoted by cokerϕ, is amorphism µ : B → C such that µϕ = 0 and if ψϕ = 0 for ψ : B → X, thenthere is a unique ψ′ : C → X such that ψ = ψ′µ. The image of a morphismϕ : A→ B, denoted by imϕ, is the kernel of the cokernel of ϕ. For example,consider Λ-left-modules A and B and the homomorphism ϕ : A→ B. Thensimply kerϕ = a ∈ A | ϕa = 0, imϕ = ϕA = b ∈ B | b = ϕa for somea ∈ A and cokerϕ = B/imϕ.

A morphism µ : A B is a monomorphism if µα = µβ implies α = β forall morphisms α, β. This lends to the dual notion. A morphism ε : A Bis an epimorphism if αε = βε implies α = β for all morphisms α, β.

3

An object P is projective if given any epimorphism Bε C, the map (P, ε) :

(P,B)→ (P,C) is surjective.

P

~~B

ε // // C

An injective object is defined dually, i.e. an object I is injective if given any

monomorphism Aι→ B, the map (ι, I) : (B, I)→ (A, I) is surjective.

A ι //

B

I

Given a family Xi of objects of a category C, the product of the Xi is(X, pi), where X is an object and the morphisms pi : X → Xi are theprojections with the universal property, this is to say, given any object Yand morphisms fi : Y → Xi, there is a unique morphism f = fi : Y → Xsuch that pif = fi. The notion of a coproduct is defined dually, i.e. givena family Xi of objects of a category C, the coproduct of the Xi is (X, qi),written X =

∐Xi, where X is an object and the morphisms qi : Xi → X are

the injections with the universal property, this is to say, given any object Yand morphisms fi : Xi → Y , there is a unique morphism f = 〈fi〉 : X → Ysuch that fqi = fi.

In the category of Λ-left-modules, the product and the coproduct are simplythe direct product and the direct sum, written

∏Xi and

⊕Xi respectively.

This reveals the notion of duality in category theory, see [18, II.3.] foran account on it. For example, a morphism ϕ is a monomorphism in acategory C if and only if ϕ is an epimorphism in Cop, and the notions aresaid to be dual to each other. Similarly, the diagram illustrating the notionof a projective object, by reversing the arrows and changing epimorphismto monomorphism, is precisely the diagram illustrating the notion of aninjective object, and vice versa, thus the notions are dual to each other, andso are the notions of a product and a coproduct. Since (Cop)op = C, anystatement involving certain notions which is true in a category C remainstrue if they are replaced by their respective dual notions, with the necessaryadjustments entailed by the changes. Though the duality pair, when itoccurs, is written in its entirety more often in the introduction chapter, theduals of the statements and proofs are usually left to the reader.

Definition 0.1.3. The finite-dimensional k-algebra Λ over the field k is saidto be right hereditary if any right ideal of Λ is projective as a Λ-module. A

4

left hereditary algebra is defined dually, i.e. the algebra Λ is said to beleft hereditary if any left ideal of Λ is projective as a Λ-module. Since thealgebra Λ is right hereditary if and only if it is left hereditary, we simply saya hereditary algebra, see the remark after [1, Theorem VII.1.4].

Let M be a Λ-right-module, N be a Λ-left-module and consider the setB = m⊗ n | m ∈M,n ∈ N, and let V be a k-vector space with B as theunderlying set. Then the tensor product of M and N , written M ⊗Λ N , isdefined to be the k-vector space V modulo the following relations.

(i) m1 ⊗ n+m2 ⊗ n = (m1 +m2)⊗ n,

(ii) m⊗ n1 +m⊗ n2 = m⊗ (n1 + n2),

(iii) mλ⊗ n = m⊗ λn, λ ∈ Λ.

The tensor product M ⊗Λ N does not necessarily have a module structure.

Lemma 0.1.4. Let M be a Λ-right-module, N be a Λ-left-module and J bea Λ-bi-module. Then

(i) M ⊗Λ Λ ∼= M and Λ⊗Λ N ∼= N ,

(ii) J ⊗Λ N is injective if J is injective as a Λ-left-module and N is pro-jective.

A category C is additive if C has a zero object, any two objects in C havea product in C and given objects A,B,C in C, the morphism sets (A,B)are abelian groups such that the composition (A,B) × (B,C) → (A,C) isbilinear. An additive category C is abelian if every morphism has a kerneland a cokernel, every monomorphism is the kernel of its cokernel and everyepimorphism is the cokernel of its kernel and finally, every morphism canbe expressed as the composite of an epimorphism and a monomorphism.The categories Mod(Λ) and mod(Λ) are examples of additive and abeliancategories, see [18, II.9.]. Finally, let F : C → D be a functor between theadditive categories. Then F is an additive functor if given X,Y in C, themap FXY : HomC(X,Y )→ HomD(FX,FY ) is a homomorphism.

A sequence A0j0→ A1 → . . .→ An

jn→ An+1 in a category C satisfies ji+1ji = 0

for 0 ≤ i ≤ n−1. A sequence Aϕ→ B

ψ→ C in Mod(Λ) is exact at B if kerψ =imϕ. A sequence A0 → A1 → . . . → An → An+1 in Mod(Λ) is exact if it isexact at all the Ai, 1 ≤ i ≤ n. The exact sequence 0 → A B C → 0in Mod(Λ) is a short exact sequence. The notion of exactness can readily begeneralized in any abelian categories, with some delicate considerations.

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A covariant functor F is left exact if given the exact sequence 0 → A →B → C, the sequence 0 → FA → FB → FC is exact. A contravariantfunctor F is left exact if given the exact sequence A → B → C → 0, thesequence 0 → FC → FB → FA is exact. The right exactness of a functorF , covariant or contravariant, is defined similarly.

The notions of a monomorphism and an epimorphism are generalizations ofthe following.

Definition 0.1.5. Let α : A→ B be a morphism in a category C. Then α isa split monomorphism if there is a morphism β : B → A such that βα = 1A.A split epimorphism is defined dually, i.e. α is a split epimorphism if thereis a morphism β : B → A such that αβ = 1B.

In a different manner, the following generalizes the notions of a monomor-phism and an epimorphism.

Definition 0.1.6. Let A and B be in a category C. A morphism α : A→ Bis right minimal if αf = α for a morphism f : A → A implies that f is anautomorphism. Dually, a morphism α : A → B is left minimal if fα = αfor a morphism f : B → B implies that f is an automorphism.

Let A be an abelian category in this chapter. The following lemma is stan-dard.

Lemma 0.1.7. (Five Lemma) In the following commutative diagram, con-sidered in A,

// A1//

ϕ1

A2//

ϕ2

A3//

ϕ3

A4//

ϕ4

A5

ϕ5

//

// B1// B2

// B3// B4

// B5// ,

assume that the rows are exact. If ϕ1, ϕ2, ϕ4 and ϕ5 are isomorphisms,then ϕ3 is also an isomorphism.

Given a category C, any functor F : C → C is an endofunctor. If there is afunctor G : C → C such that FG = GF = id, then F is an automorphismand written G = F−1. Given functors F,G : C → D and objects X, Yin C, a natural transformation t from F to G is a collection of morphismstX : FX → GX in D such that for any morphism f : X → Y in C, thereis the relation G(f)tX = tY F (f), as shown in the following commutativediagram.

FXtX //

Ff

GX

Gf

FYtY // GY

6

If tX is an isomorphism for each X, then t is a natural equivalence andwe write F ' G. Let F : C → D, G : D → C be functors such thatGF ' id : C → C and FG ' id : D → D. Then F (and also G) is anequivalence and they are quasi-inverses of each other, and the categoriesC and D are equivalent categories. If C = D, then F (and also G) is anautoequivalence. For example, an automorphism is an antoequivalence.

Example 0.1.8. (i) ([18, Exercise II.4.1.]) Let C0 be a full subcategory ofC such that given any object A in C, there is precisely one object A0

in C0 with A0∼= A. Then C0 is equivalent to C, and the subcategory

C0 is said to be a skeleton of C. Along with this, any two skeletons ofC are isomorphic.

(ii) ([31, Theorem IV.4.1]) Consider the functor F : C → D. Then F is anatural equivalence of categories if and only if F is full, faithful andessentially surjective, this is to say, for all Y in D, there is some X inC such that FX ∼= Y .

The object X in an additive category C is indecomposable if X is non-zeroand if X ∼= X1 ⊕ X2, then either X1

∼= 0 or X2∼= 0. If C is the module

category Mod(Λ), then End(X) is also a k-algebra, where End(X) is theendomorphism ring Hom(X,X). A k-algebra A is a local algebra if A hasa unique maximal right ideal. A characterization of a finite-dimensionalk-algebra A as a local algebra is given in [1, Lemma I.4.6].

Lemma 0.1.9. ([1, Corollary I.4.8]) Let M be a Λ-left-module. Then M isindecomposable if and only if the algebra End(M) is local.

Let k be a field and let X,Y, Z be objects in a category C. Then C is saidto be k-linear if it is additive, each Hom set (X,Y ) is a k-vector space andeach composition map (X,Y )× (Y, Z)→ (X,Z) is bilinear. It is Hom finiteif given any X,Y in C, the morphism space (X,Y ) is a finite-dimensionalk-vector space.

A category C is finite-dimensional k-additive if C is k-linear and all Hom sets(X,Y ) are finite-dimensional k-vector spaces. If all idempotents split, thatis, given e = e2 in (X,X) for an object X in C, there are maps µ : Y → Xand ρ : X → Y with ρµ = 1Y and µρ = e, then the endomorphism ringEnd(X) of any indecomposable object X of C is a local ring, so that thecategory C is to be a Krull-Schmidt category, see the following definition in[41, 2.2].

Definition 0.1.10. (c.f. Example 0.1.8) Let C be a finite-dimensional k-linear category. Then C is Krull-Schmidt if given indecomposable objectsxi, xj , 1 ≤ i ≤ s, 1 ≤ j ≤ t, in C such that

⊕si=1 xi

∼=⊕t

j=1 yj , then s = tand there is a permutation π of 1, . . . , s such that xi ∼= yπ(i) for all i.

7

For example, the category mod(Λ) is Krull-Schmidt, where we remind thereader that Λ is a finite-dimensional k-algebra over the field k.

Notation 0.1.11. Let C be a Krull-Schmidt category, and let Xi be a com-plete set of representatives from the isomorphism classes of indecomposableobjects in C. The full subcategory of C, with objects the Xi, is denoted byindC. Let C be any additive category and C0 a subcategory of C. Thendenote by addC0 the smallest subcategory of C containing all the objects inC0, which is closed under direct sums and direct summands. The categoryaddC0 is said to be the additive closure of C0, or informally, the add ofobjects in C0.

Definition 0.1.12. Let C be an additive category. A chain complex C =Cn, ∂n over C is

C = · · · // Cn+1∂n+1 // Cn

∂n // Cn−1∂n−1 // Cn−2

// · · · ,

where the Ci are in C and ∂i∂i+1 = 0.

The following is the dual notion.

A cochain complex C = Cn, ∂n over C is

C = · · · // Cn−2 ∂n−2// Cn−1 ∂

n−1// Cn

∂n // Cn+1 // · · · ,

where the Ci are in C and ∂i+1∂i = 0.

Remark 0.1.13. Contexts written in terms of chain complexes can be trans-lated suitably in the context of cochain complexes, where the prefix “co-” isto be added when it makes sense, and vice versa. This is to be seen read-ily as subscripts are used for chain complexes and superscripts for cochaincomplexes.

A chain complex C = Cn, ∂n is said to be bounded below if Cn = 0 forn 0. Similarly, there is the notion of a chain complex which is boundedabove or bounded on both sides. An object M in C is identified with thechain complex

· · · // 0 // 0 //M // 0 // 0 // · · ·

with M in degree 0 when there is no confusion.

Definition 0.1.14. Let C be an additive category and let C = Cn, ∂n andD = Dn, ∂n be chain complexes over C. Then a chain map ϕ : C → D isa family ϕn : Cn → Dn of morphisms such that the following diagram,

Cn∂n //

ϕn

Cn−1

ϕn−1

Dn

∂n // Dn−1,

8

is commutative for each value of n.

Given an additive category C, the category with (co)chain complexes overC as objects and (co)chain maps as morphisms is denoted by C(C). Thecategory of (co)chain complexes over Mod(Λ) is denoted by C(Λ).

Definition 0.1.15. Let C be an additive category and let C = Cn, ∂n andD = Dn, ∂n be chain complexes over C. Given a chain map ϕ : C → D, let

the mapping cone of ϕ be the chain complex E = E(ϕ) = En, ˜∂n, where

En = Cn−1 ⊕ Dn and˜∂n(a, b) = (−∂n−1a, ϕn−1a + ∂nb) for a ∈ Cn−1 and

b ∈ Dn.

Definition 0.1.16. Let C = Cn, ∂n be a chain complex in C(Λ). LetHn(C) = ker∂n/im∂n+1 be the n-th homology module of C and write H(C) =Hn(C). Also Zn = Zn(C) = ker∂n.

Let C = Cn, ∂n and D = Dn, ∂n be chain complexes in C(Λ) andconsider the chain map ϕ : C → D. Then it is natural to define Hn(ϕ) :Hn(C) → Hn(D) to be Hn(ϕ)(xn + im∂n+1) = ϕn(xn) + im∂n+1 for xn inker∂n. Since xn is in ker∂n and ∂nϕn(xn) = ϕn−1∂n(xn) = 0, thereforeϕn(xn) is indeed in ker∂n.

The map Hn(ϕ) is well-defined. Consider the following diagram,

Cn+1∂n+1 //

ϕn+1

Cn∂n //

ϕn

Cn−1

ϕn−1

Dn+1

∂n+1 // Dn∂n // Dn−1.

Let xn and x′n be in ker∂n such that xn + im∂n+1 = x′n + im∂n+1, whichis to say xn − x′n is in im∂n+1. Therefore xn − x′n = ∂n+1(u) for some uin Cn+1. Hence ϕnxn − ϕnx′n = ϕn(xn − x′n) = ϕn∂n+1(u) = ∂n+1ϕn+1(u),which gives ϕnxn − ϕnx′n in im∂n+1 indeed.

Example 0.1.17. Functors can be defined on different levels. Here are a fewexamples.

(i) A group can be viewed as a category with a single object. The mor-phisms of the category are the elements of the group, which are allinvertible. A functor between two groups is then a homomorphismbetween them.

(ii) Consider the chain complex X = Xn, ∂n. The suspension functorΣ is defined by shifting the complex X one place to the left, andchanging the sign of the differential ∂, i.e. for the chain complexΣ(X) = Σ(X)n, ∂n, let Σ(X)n = Xn−1 and ∂n = −∂n−1.

9

(iii) Let C and D be two chain complexes in C(Λ) with a chain map ϕ :C → D. This induces a well-defined morphism H(ϕ) : H(C) → H(D)and the homology functor H(−) correspondingly, by the little descrip-tion after Definition 0.1.16. The chain map ϕ is a quasi-isomorphismif the induced homomorphisms Hn(ϕ) are isomorphisms.

(iv) Let A and B be arbitrary categories. A pair of functors L : A → Band R : B → A are adjoints if there is an isomorphism τ : (LA,B)

∼=→(A,RB) for all A in A and B in B, such that τ is natural in A andB. For the isomorphism τ to be natural in A is to say that given amorphism f : A1 → A2 in A, the following diagram is commutative.

(LA2, B)τ //

(A2, RB)

(LA1, B)

τ // (A1, RB)

Similarly, for the isomorphism τ to be natural in B is to say that givena morphism g : B1 → B2 in B, the following diagram is commutative.

(LA,B1)τ //

(A,RB1)

(LA,B2)

τ // (A,RB2)

L is the left adjoint and R is the right adjoint of the pair of functors.

In Example 0.1.17(iv), givenA inA andB in B, the map ηA = τ(idLA) : A→RLA is the unit of the adjunction, and the map εB = τ−1(idRB) : LRB → Bis the counit of the adjunction. If A and B are abelian, then L is right exactand R is left exact, see [42, Theorem 2.14]. In this way, adjoints approximateexactness as well.

Intuitively, adjoints are approximations of inverse functors. Indeed supposethe functors L and R are inverse functors. Then their actions can be re-trieved through τ . For example, let A1 and A2 be in A and consider themorphism f : A1 → A2. Suppose A2 = RB2 for some B2 in B. Then themorphism Lf : LA1 → LA2, where LA2 = LRB2 = B2, takes the valueτ−1f .

Any two left adjoints of a functor R are naturally equivalent, dually, anytwo right adjoints of a functor L are naturally equivalent, see [11, A5.2.1].A functor with a left adjoint respects products, and a functor with a rightadjoint respects coproducts. Any invertible functor has both a right and aleft adjoint, see [35, Proposition 1.1.6].

The following lemma follows from the definition of a quasi-isomorphism.

10

Lemma 0.1.18. Let X be a chain complex in C(Λ) and consider the mor-phism f : 0→ X in C(Λ). Then f is a quasi-isomorphism if and only if Xis exact.

Lemma 0.1.19. ([18, Theorem IV.2.1]) Let C be an abelian category andlet A B C be a short exact sequence of chain complexes over C, whichis equivalent to saying An Bn Cn is a short exact sequence in C foreach value of n. Then there is a long exact sequence

· · · // Hn(A) // Hn(B) // Hn(C) // Hn−1(A) // · · ·

in C.

Lemma 0.1.20. Consider a chain map ϕ : C → D in C(Λ) and the mappingcone E = E(ϕ) in Definition 0.1.15. Let Σ be the suspension functor. Then

there is a short exact sequence Dı E(ϕ)

ρ ΣC in C(Λ), where ı is the

canonical inclusion and ρ is the canonical surjection. In addition,

(i) By Lemma 0.1.19, there is a long exact sequence

· · · // Hn(C) // Hn(D) // Hn(E(ϕ)) // Hn−1(C) // · · ·

in Mod(Λ). Therefore H(E(ϕ)) = 0 if and only if H(ϕ) : H(C)∼=→

H(D).

(ii) Let X be a chain complex in C(Λ) and consider the chain map µ : X →E(ϕ) such that ρµ = 0. Then there is a unique chain map µ′ : X → Dsuch that ıµ′ = µ.

X

µ

0

##

µ′

||0 // D

ı // E(ϕ)ρ // ΣC // 0

In Lemma 0.1.20(i), the map ϕ is a quasi-isomorphism if and only if themapping cone E = E(ϕ) is exact.

The construction of a projective resolution is a subtle way of encoding amodule A in Mod(Λ) into a complex in C(Λ).

Definition 0.1.21. Let A be in Mod(Λ). A projective resolution P (A) ofA is a chain complex

P (A) = · · · // P2// P1

// P0// 0 // · · ·

in C(Λ) where Pi is projective for all i ≥ 0, Hi(P ) = 0 for all i ≥ 1 andH0(P ) ∼= A.

11

Definition 0.1.22. Let A be in Mod(Λ). An injective resolution I(A) of Ais a cochain complex

I(A) = · · · // 0 // I0 // I1 // I2 // · · ·

in C(Λ) where Ii is injective for all i ≥ 0, Hi(I) = 0 for all i ≥ 1 andH0(I) ∼= A.

Example 0.1.23. Below are some examples of projective resolutions.

(i) Let k be a field. Consider the ring R = k[X]/(X2), and turn k intoan R-module by letting (k0 + k1X)m = k0m for m in k. Then theprojective resolution P (M) of k is

P (M) = · · · // k[X]/(X2)·X // k[X]/(X2) // 0 // · · · .

(ii) Let k be a field. Consider the ring R = k[X], and turn k into anR-module by letting (k0 + k1X + . . . + knX

n)m = k0m for m in k.Then the projective resolution P (M) of k is

P (M) = · · · // 0 // k[X]·X // k[X] // 0 // · · · .

Definition 0.1.24. ([18, VII.7.]) The global dimension of Λ, denoted by gl.dim. Λ, is less than or equal to m if for all Λ-modules A and all projectiveresolutions P (A) of A, ker(Pm−1 → Pm−2) is projective. Then gl. dim. Λ= m if gl. dim. Λ ≤ m but gl. dim. Λ m− 1.

Example 0.1.25. ([1, Theorem VII.1.4]) The algebra Λ is hereditary if andonly if the global dimension of Λ is at most one.

Definition 0.1.26. Let C be an additive category and let C = Cn, ∂n andD = Dn, ∂n be chain complexes over C. A homotopy h : ϕ → ψ betweenchain maps ϕ,ψ : C → D is a collection of morphisms hn : Cn → Dn+1such that ϕn − ψn = ∂n+1hn + hn−1∂n.

C = · · · //Cn+1//Cn

∂n //

hn

||

Cn−1//

hn−1

||

· · ·

D = · · · //Dn+1∂n+1 //Dn

//Dn−1// · · ·

If there is a homotopy h : ϕ → ψ, then ϕ,ψ are homotopic and we writeϕ ' ψ. The homotopy relation is an equivalence relation, so that given achain map ϕ, let ϕ be the equivalence class of ϕ. The chain complexes Cand D are homotopy equivalent if there are chain maps ϕ : C → D andψ : D → C such that ψϕ ' id and ϕψ ' id.

12

A chain complex C in C(Λ) is projective if Cn is projective for all n ≥ 0 andCn = 0 for all n ≤ −1. A cochain complex D in C(Λ) is injective if Dn isinjective for all n ≥ 0 and Dn = 0 for all n ≤ −1. A (co)chain complex Cin C(Λ) is acyclic if (Hn(C) = 0) Hn(C) = 0 for n ≥ 1.

Lemma 0.1.27. (i) ([18, Theorem IV.4.1]) Let C be a projective chaincomplex and D be an acyclic chain complex in C(Λ). Then for everyhomomorphism ϕ : H0(C)→ H0(D), there is a chain map ϕ inducingϕ and any two such chain maps are homotopic.

(ii) ([18, Theorem IV.4.4]) Let C be an acyclic cochain complex and D bean injective cochain complex in C(Λ). Then for every homomorphismϕ : H0(C) → H0(D), there is a chain map ϕ inducing ϕ and any twosuch chain maps are homotopic.

Example 0.1.28. For any M in Mod(Λ), the projective resolution P (M)always exists, see [18, Lemma IV.4.2]. Consider the isomorphism f0 :

H0P (M)∼=→ M . By Lemma 0.1.27(i), the morphism f0 can be extended

to a chain map f : P (M) → M . The chain map f is a quasi-isomorphism.It follows from Lemma 0.1.27(i) that any two projective resolutions of Mare homotopy equivalent.

Lemma 0.1.29. Let P be a chain complex in C(Λ). If P is projective, thenH(P ) = 0 if and only if 1 ' 0 : P → P .

Proof. (if) Since 1 ' 0 : P → P , there are morphisms hn : Pn → Pn+1 suchthat 1 = ∂n+1hn + hn−1∂n.

P = · · · //Pn+1//Pn

∂n //

hn

||

Pn−1//

hn−1

||

· · · //P0

P = · · · //Pn+1∂n+1 //Pn //Pn−1

// · · · //P0

Let y ∈ ker∂n. Therefore y = (∂n+1hn + hn−1∂n)y = (∂n+1hn)y and soy ∈ im∂n+1. (only if) Consider the morphism π : H0(P ) → H0(P ). SinceH0(P ) = 0, therefore π = 1 = 0. Lemma 0.1.27(i) then gives 1 ' 0 : P →P .

Lemma 0.1.30. ([18, Exercise IV.3.1.]) Let C and D be chain complexesin C(Λ) and let ϕ and ψ be homotopic chain maps : C −→ D. Then themapping cones E(ϕ) and E(ψ) are isomorphic.

The homotopy category K(Λ) is a category whose objects are the chaincomplexes in C(Λ), and whose morphisms are the homotopy equivalence

13

classes of chain maps between chain complexes, see [17, Chapter I. §2.].This is possible because the null homotopic morphisms form an ideal inC(Λ). Let K+(Λ), K−(Λ) and Kb(Λ) be the full subcategories of K(Λ)consisting of the complexes bounded below, bounded above and boundedon both sides, respectively.

Lemma 0.1.31. Let X and Y be chain complexes in K(Λ) and let ϕ and ψbe homotopic chain maps : X → Y . Let F : K(Λ) → K(Λ) be an additivefunctor and consider the homology functor H(−). Then

(i) H(ϕ) = H(ψ),

(ii) Fϕ ' Fψ,

(iii) The morphism f : 0 → X in K(Λ) is an isomorphism if and only ifX is null homotopic, i.e. X ' 0.

Proof. This is trivial.

By virtue of Lemma 0.1.31, the homology functor H(−) and subsequentlyquasi-isomorphisms are well-defined on K(Λ).

Lemma 0.1.32. Let P be a projective module in Mod(Λ) and let X be inC(Λ). Then HomΛ(P,H0X) ∼= H0(HomΛ(P,X)).

Proof. Let the chain complex X = Xn, ∂n be

X = · · · // X2// X1

∂1 // X0∂0 // X−1

// X−2// · · · ,

and the chain complex (P,X) be

(P,X) = · · · // (P,X1)d1 // (P,X0)

d0 // (P,X−1) // · · · ,

where di = (P, ∂i).

Let θ : HomΛ(P,H0X) → H0(HomΛ(P,X)) and θ′ : H0(HomΛ(P,X)) →HomΛ(P,H0X) be maps between the two Hom sets, which are going to bedefined. Let p be in P . Given a morphism f : P → H0X where f(p) = xp +im∂1 for some xp in ker∂0, consider the canonical morphism πX : ker∂0 H0(X). Since P is projective, there is a morphism g : P → ker∂0 such thatπXg = f .

ker∂0πX // // H0(X)

P

g

OO

f

::

14

This gives a morphism g : P → X0. Since ∂0g = 0, therefore g is in kerd0

and define θ(f) = g + imd1 in H0(HomΛ(P,X)). Conversely, consider themorphism g + imd1 in H0(HomΛ(P,X)) where g : P → X0 and ∂0g = 0.Given p in P , define θ′(g + imd1) = f where f(p) = g(p) + im∂1. Finally,one can verify that θ and θ′ are well-defined and that θθ′ = θ′θ = 1.

0.2 Triangulated categories

This section is in reminiscence of [17] and [44], though the reader can alsoseek guidance from [24], [34], [35] and [25].

Unlike an abelian category, whose properties are inherent in the category, atriangulated category is a category with extra structure layered on it. Letus begin with the definition.

Definition 0.2.1. ([17, Chapter I. §1.]) A triangulated category is an addi-tive category T, together with

(a) an automorphism Σ : T→ T known as the translation functor, and

(b) a collection of sextuples (x, y, z, u, v, w) known as the distinguishedtriangles of T, written x

u→ yv→ z

w→ Σx, or simply xu→ y

v→ z →.Here x, y, z are objects and u, v, w are morphisms of T.

A morphism from the distinguished triangle (x, y, z, u, v, w) to the distin-guished triangle (x′, y′, z′, u′, v′, w′) is a commutative diagram,

xu //

f

yv //

g

zw //

h

Σx

Σf

x′u′ // y′

v′ // z′w′ // Σx′.

If the morphisms f , g and h are isomorphisms, the distinguished triangles(x, y, z, u, v, w) and (x′, y′, z′, u′, v′, w′) are then said to be isomorphic.

The above is required to satisfy

(i) (TR1) Every sextuple (x, y, z, u, v, w) isomorphic to a distinguishedtriangle is a distinguished triangle. Every morphism u : x→ y can beimbedded in a distinguished triangle (x, y, z, u, v, w). For each x, thesextuple (x, x, 0, idx, 0, 0) is a distinguished triangle.

15

(ii) (TR2) (x, y, z, u, v, w) is a distinguished triangle if and only if (y, z,Σ(x), v, w,−Σ(u))is a distinguished triangle.

(iii) (TR3) Given distinguished triangles (x, y, z, u, v, w) and (x′, y′, z′, u′, v′, w′),and morphisms f and g such that gu = u′f , there is a morphism hsuch that the following diagram is commutative.

xu //

f

yv //

g

zw //

h

Σx

Σf

x′u′ // y′

v′ // z′w′ // Σx′

(iv) (TR4) The octahedral axiom. Given two composable morphisms f :x→ y and g : y → u, as in the following commutative diagram,

xf //

gf

y

g

u u,

we can extend it to the following commutative diagram,

xf //

gf

y

g

// z

// Σx

u

u //

0

// Σu

v //

w

// Σz // Σv

Σx // Σy // Σz // Σ2x,

where all the rows and columns are distinguished triangles.

The automorphism Σ : T → T is an additive functor. By abuse of nota-tion, given a distinguished triangle x

u→ yv→ z

w→ Σx, the object z is themapping cone of the morphism u : x→ y, see Definition 0.1.15. By (TR1),any object x in a triangulated category T yields a distinguished triangle(x, x, 0, idx, 0, 0). This is similar to the way any object in a category yieldsa chain complex, see Definition 0.1.12.

The following is a slight variation of (TR3).

(TR3’) Given distinguished triangles (x, y, z, u, v, w) and (x′, y′, z′, u′, v′, w′),and morphisms f and h such that Σf w = w′h, there is a morphism g such

16

that the following diagram is commutative.

xu //

f

yv //

g

zw //

h

Σx

Σf

x′u′ // y′

v′ // z′w′ // Σx′

This follows from (TR2) and (TR3).

A full additive subcategory S in T is a triangulated subcategory if it is closedunder isomorphisms and the translation functor Σ, and for any distinguishedtriangle x→ y → z → Σx, if x and y are in S, then z is in S. Let T and T′

be triangulated categories with translation functors Σ and Σ′ respectively.A triangulated functor from T to T′ is a pair (F, σ) where F : T → T′ is

an additive functor and σ : FΣ'→ Σ′F , such that given a distinguished

triangle xα→ y

β→ zγ→ Σx in T, the image Fx

Fα→ FyFβ→ Fz

σxFγ→ Σ′Fxis a distinguished triangle in T′. If a triangulated functor F : T → T′ isan equivalence, then F is an triangle equivalence and T and T′ are triangleequivalent.

Lemma 0.2.2. (i) Let xα→ y

β→ z → Σx be a distinguished triangle.Then the composition βα is zero.

(ii) In (TR3), if the morphisms f and g are isomorphisms, then the mor-phism h is an isomorphism.

(iii) Let xα→ y → z → Σx be a distinguished triangle. Then α is an

isomorphism if and only if z ∼= 0.

(iv) If x1 → y1 → z1 → Σx1 and x2 → y2 → z2 → Σx2 are distinguishedtriangles, then x1 ⊕ x2 → y1 ⊕ y2 → z1 ⊕ z2 → Σx1 ⊕ Σx2 is a

distinguished triangle. In particular, x → x ⊕ z → z0→ Σx is a

distinguished triangle.

(v) The distinguished triangles x→ y → zγ→ Σx and x→ x⊕z → z

0→ Σxare isomorphic if and only if γ = 0.

(vi) Let xα→ y

β→ z → Σx be a distinguished triangle. If there is amorphism f : m → y such that βf = 0, then there is a morphismf ′ : m→ x such that αf ′ = f .

m

f

0

f ′

x

α // yβ // z // Σx

17

Proof. (i) and (ii): This is [17, Proposition I.1.1.]. (iii) This is [35, Corollary1.2.6]. (iv) This is [35, Proposition 1.2.3]. (v) This is [35, Corollary 1.2.7].

(vi) By (TR1) and (TR2), there is a distinguished triangle 0→ mid→ m→ 0,

and by (TR2), there is a distinguished triangle Σ−1z → xα→ y

β→ z. By(TR3’), there is a morphism f ′ : m→ x such that the following diagram iscommutative.

0 //

mid //

f ′

m //

f

0

0

Σ−1z // x

α // yβ // z

This gives αf ′ = f .

Lemma 0.2.2(vi) suggests that a triangulated category might not permitkernels, since the morphism f ′ is not unique. A distinguished triangle of the

form x→ x⊕ z → z0→ Σx is said to be a split distinguished triangle.

Lemma 0.2.3. ([14, Remark 3.2]) Let xα→ y

β→ zγ→ Σx be a distinguished

triangle. Then the following are equivalent.

(i) γ 6= 0,

(ii) α is not a split monomorphism,

(iii) β is not a split epimorphism.

Lemma 0.2.4. ([30, Lemma 2.5]) Let xα→ y

β→ zγ→ Σx be a distinguished

triangle. Then β is right minimal if and only if α is left minimal.

In each of the next three little sections a different flavour of triangulated cat-egories is introduced. They show how triangulation is distilled and shapedin different categories.

0.2.1 The homotopy category

The homotopy category K(Λ) is triangulated, see [17, Chapter I. §2.]. Thisis described as follows. Let the translation functor Σ take the value of thesuspension functor. Given a morphism ϕ : X → Y in K(Λ), complete it to a

diagram Xϕ→ Y → E(ϕ)→ ΣX, where E(ϕ) is the mapping cone and the

morphisms are viewed in K(Λ). By Lemma 0.1.30, the mapping cone E(ϕ)is independent of the choice of ϕ up to isomorphism. Let the distinguishedtriangles in K(Λ) be either diagrams of such a form or diagrams isomorphicto them.

18

The following lemma is essentially a verification of (TR2) in the categoryK(Λ).

Lemma 0.2.5. Consider the identity chain map 1X : X → X in C(Λ).Then the mapping cone E(1X) is isomorphic to 0 in K(Λ).

Proof. For (xn−1, xn) in Xn−1 ⊕Xn, define hn : Xn−1 ⊕Xn → Xn ⊕Xn+1

to be hn(xn−1, xn) = (xn, 0). Consider the following diagram,

E(1X) = · · · // Xn+1 ⊕Xn+2∂n+2 // Xn ⊕Xn+1

∂n+1 //

hn+1

vv

Xn−1 ⊕Xn∂n //

hn

ww

· · ·

E(1X) = · · · // Xn+1 ⊕Xn+2∂n+2 // Xn ⊕Xn+1

∂n+1 // Xn−1 ⊕Xn∂n // · · · .

Since (∂n+2hn+1+hn∂n+1)(xn, xn+1) = (∂n+2hn+1)(xn, xn+1)+(hn∂n+1)(xn, xn+1) =∂n+2(xn+1, 0) + hn(−∂nxn, xn + ∂n+1xn+1) = (−∂n+1xn+1, xn+1) + (xn +∂n+1xn+1, 0) = (xn, xn+1), therefore ∂n+2hn+1 +hn∂n+1 = 1. Consequently,the identity chain map on E(1X) is homotopic to 0, and so HomK(Λ)(E(1X), E(1X)) =0 whence E(1X) ∼= 0 in K(Λ).

Lemma 0.2.6. Let P be a projective module in Mod(Λ) and let X be inK(Λ). Then HomK(Λ)(P,X) ∼= HomΛ(P,H0(X)).

Proof. Let the chain complex X = Xn, ∂n be

X = · · · // X2// X1

∂1 // X0∂0 // X−1

// X−2// · · · .

Let θ : HomK(Λ)(P,X) → HomΛ(P,H0(X)) and θ′ : HomΛ(P,H0(X)) →HomK(Λ)(P,X) be maps between the two Hom sets, which are going tobe defined. Consider a morphism f in HomΛ(P,H0(X)) and the canonicalmorphism πX : ker∂0 H0(X). Since P is projective, there is a morphismf ′′0 : P → ker∂0 such that πXf

′′0 = f .

ker∂0πX // // H0(X)

P

f ′′0

OO

f

::

Extend f ′′0 to a morphism f ′0 : P → X0, which can readily be extended toa chain map f ′ in HomC(Λ)(P,X) and subsequently to a morphism f ′ in

HomK(Λ)(P,X) so that we define θ′(f) = f ′.

19

P = · · · // 0 //

0 //

P //

f ′0

0 //

0 //

· · ·

X = · · · // X2∂2 // X1

∂1 // X0∂0 // X−1

∂−1 // X−2// · · ·

Conversely, suppose a morphism g in HomK(Λ)(P,X) is given. In particular,consider the morphism g0 : P → X0 obtained from the chain map g inHomC(Λ)(P,X).

P = · · · // 0 //

0 //

P //

g0

0 //

0 //

· · ·

X = · · · // X2∂2 // X1

∂1 // X0∂0 // X−1

∂−1 // X−2// · · ·

Let us define θ(g) = h where h(p) = g0(p) + im∂1 for p in P . Finally, onecan verify that θ and θ′ are well-defined and that θθ′ = θ′θ = 1.

Example 0.2.7. Below is a special case in Lemma 0.2.6. Let P be the pro-jective module Λ and consider the following diagram.

P = · · · // 0 //

0 //

Λ //

0 //

||

0 //

· · ·

X = · · · // X2∂2 // X1

∂1 // X0∂0 // X−1

∂−1 // X−2// · · ·

Then HomK(Λ)(P,X) ∼= ker∂0/im∂1 = H0(X) ∼= HomΛ(P,H0(X)).

The following lemma is a little consequence of Lemma 0.2.6.

Lemma 0.2.8. Let P be a projective module in Mod(Λ) and let α : A→ B

be a quasi-isomorphism in K(Λ). Then HomK(Λ)(P, α) : HomK(Λ)(P,A)∼=→

HomK(Λ)(P,B).

Proof. Consider the chain complexes A = An, ∂An , B = Bn, ∂Bn and thequasi-isomorphism α : A→ B in the following diagram.

A = · · · // A2∂A2 //

α2

A1∂A1 //

α1

A0∂A0 //

α0

A−1

∂A−1 //

α−1

A−2//

α−2

· · ·

P

g0

??

h0

B = · · · // B2

∂B2 // B1∂B1 // B0

∂B0 // B−1

∂B−1 // B−2// · · ·

20

The lemma is to say that given a morphism h : P → B in HomK(Λ)(P,B),

there is a unique morphism g : P → A such that α g = h. One can see theexistence of g, and here we shall only show its uniqueness.

Suppose there are morphisms g and g′ in HomK(Λ)(P,A) such that α g = α

g′ = h in HomK(Λ)(P,B). By Lemma 0.2.6, there is an isomorphism θ :

HomK(Λ)(P,B)∼=→ HomΛ(P,H0(B)). Hence θ( α g) = θ( α g′), where θ( α

g) = f with f(p) = α0g0(p) + im∂B1 and where θ( α g′) = f ′ with f ′(p) =α0g

′0(p) + im∂B1 for p in P . Hence α0g0(p) + im∂B1 = α0g

′0(p) + im∂B1 , which

gives g0(p)+im∂A1 = g′0(p)+im∂A1 , so that g = g′ again by Lemma 0.2.6.

The following lemma is a triangulated version of Lemma 0.1.19.

Lemma 0.2.9. ([17, Proposition I.1.1.]) Given a distinguished triangle(X,Y, Z, u, v, w) and an object M in K(Λ), there is a long exact sequence

· · · // (M,Σ−1Z) // (M,X) // (M,Y ) // (M,Z) // (M,ΣX) // · · ·

in Mod(Z), where all the Hom groups are considered in K(Λ). In particular,if

M = · · · // 0 // 0 // Λ // 0 // 0 // · · · ,

then there is the long exact sequence

· · · // H1(Z) // H0(X) // H0(Y ) // H0(Z) // H−1(X) // · · ·

by Lemma 0.2.6.

In Lemma 0.2.9, the functor (M,−) is said to be a homological functor. Thefunctor (−,M) is also a homological functor.

Corollary 0.2.10. Let Xf→ Y → Z → be a distinguished triangle in K(Λ).

Then by Lemma 0.2.9, f is a quasi-isomorphism if and only if Z is exact.

Lemma 0.2.11. ([29, Proposition 2.5.3]) Let Xf→ Y → Z → be a distin-

guished triangle in K(Λ). Then Z is quasi-isomorphic to the mapping coneE(f).

0.2.2 The derived category

This section introduces one example of a quotient category. A collection Sof morphisms in a category C is a multiplicative system if it satisfies thefollowing axioms, see [17, Chapter I. §3.].

21

(i) (FR1) If f, g are in S, then fg is in S. For any object X in C, idX isin S.

(ii) (FR2) Any diagram Z

s

Xu // Y

with s in S can be completed to a commutative diagram,

Wv //

t

Z

s

Xu // Y,

with t in S. Similarly for the opposite statement.

(iii) (FR3) If f, g : X −→ Y are morphisms in C, then the following condi-tions are equivalent.

(i) There is an s : Y −→ Y ′ in S such that sf = sg,

(ii) There is a t : X −→ X ′ in S such that ft = gt.

The category of fractions S−1C is the localization of C with respect to S.Namely, S−1C has the same objects as C, and a morphism from X to Y inS−1C is represented by the following diagram,

Zs

f

X Y,

where s : Z → X is a morphism in S and f : Z → Y is a morphism inC. The equivalence relation on and composition of morphisms in S−1C aregiven in [17, Chapter I. §3.], and are reproduced here.

(i) (Equivalence relation on morphisms) Given morphisms from X to Yin S−1C represented by the following diagrams,

X ′

s

~~

a

X Y

22

andX ′′

t

~~

b

X Y,

the diagrams represent the same morphism if and only if there is amorphism u : X ′′′ → X in S and morphisms f : X ′′′ → X ′, g : X ′′′ →X ′′ in C such that sf = u = tg and af = bg, as shown in the followingdiagram.

X ′′′

f

g

""X ′

s

~~

a

**

X ′′t

tt

b

!!X Y

(ii) (Composition of morphisms) To compose morphisms

X ′

s

~~

a

X Y

and

Y ′

t

b

Y Z,

consider the following commutative diagram given by (FR2),

X ′′

t′

c

!!X ′

s

~~

a

!!

Y ′

t

b

X Y Z,

and then takeX ′′

st′

bc

X Z

23

to be the composition.

The canonical functor π : C → S−1C maps a morphism f : X → Y in C tothe morphism

Xid

~~

f

X Y

in S−1C.

Lemma 0.2.12. Let C be a category and S be a multiplicative system in C.Consider the canonical functor π : C→ S−1C. Then

(i) Given a morphism s : X → Y in S, the morphism π(s) is invertible.

(ii) If S consists of isomorphisms, then π is an equivalence of categories.

Proof.

(i) By definition, π(s) =

Xid

~~

s

X Y,

and then naturally π(s)−1 =

Xs

~~

id

Y X.

(ii) Consider the canonical functor π : C→ S−1C and a morphism f : X → Yin C. Then π(f) =

Xid

~~

f

X Y.

Also define the functor G : S−1C→ C by

G

Zt

~~

d

X Y

= dt−1 : X → Y .

Accordingly, G is well-defined and is a quasi-inverse to F = π.

24

Now let D be a category and consider a functor F : C→ D. Suppose F (s) isan isomorphism for any s in S. Then there is a unique functor F : S−1C→ Dsuch that the following diagram commutes, i.e. F π = F .

Cπ //

F

S−1C

F||D

This is the universal property of S−1C, see [17, Chapter I. §3.].

Let C be a triangulated category with translation functor Σ, and let S be amultiplicative system in C such that, other than (FR1) - (FR3), the followingtwo axioms are also satisfied.

(iv) (FR4) s is in S if and only if Σ(s) is in S,

(v) (FR5) The same as (TR3), but where f, g are assumed to be in S, andh is required to be in S.

The multiplicative system S is then said to be compatible with the triangu-lation, see [17, Chapter I. §3.]. Then by [17, Proposition I.3.2], the categoryS−1C has a unique triangulated structure such that the canonical functorπ : C→ S−1C is triangulated. Given a morphism from X to Y in S−1C,

Zs

f

X Y,

complete the morphism f : Z → Y in C to a distinguished triangle Zf→ Y →

C → ΣZ in C. Then take πXθ→ πY → πC → ΣπX, where θ = (πf)(πs)−1,

to be a distinguished triangle in S−1C.

Definition 0.2.13. Let C be a category and S be a multiplicative systemin C. An object P in C is K-projective if given a morphism s : X → Y in S,the map (P, s) : (P,X)→ (P, Y ) is bijective.

Lemma 0.2.14. Let C be a category and S be a multiplicative system in C.Let P in C be K-projective. Then HomS−1C(P, Y ) ∼= HomC(P, Y ) for any Yin C.

Proof. Let θ : HomS−1C(P, Y ) → HomC(P, Y ) and θ′ : HomC(P, Y ) →HomS−1C(P, Y ) be maps between the two Hom sets, which are going to be

25

defined. Given a morphism g : P → Y in C, define θ′(g) to be the followingmorphism

Pid

g

P Y

in S−1C. Conversely, given the following morphism µ =

Zs

f

P Y

in HomS−1C(P, Y ), where f : Z → Y is any morphism in C, there is the

bijective map (P, s) : (P,Z)∼=→ (P, P ), since s : Z → P is a morphism

in S and P is K-projective. In particular, consider the identity morphismid : P → P in (P, P ). Then there is a unique p in (P,Z) such that sp = id,and θ(µ) is defined to be fp in HomC(P, Y ).

Finally, one can verify that θ and θ′ are well-defined and that θθ′ = 1. Onthe other hand, since the diagrams

Zs

f

P Y

and

Psp=id

fp

P Y

represent the same morphism in HomS−1C(P, Y ), therefore θ′θ does indeedequal 1.

Consider the category K(Λ). By [17, Proposition I.4.1], the class of quasi-isomorphisms forms a multiplicative system S in K(Λ) compatible with thetriangulation. The category of fractions S−1K(Λ) is defined to be the derivedcategory D(Λ) of Λ. Similarly, let S+ be the class of quasi-isomorphisms in-side K+(Λ). Then S+ forms a multiplicative system in K+(Λ), and we defineD+(Λ) = (S+)−1K+(Λ), D−(Λ) and Db(Λ). They are full subcategories ofD(Λ), and D+(Λ) ∩D−(Λ) = Db(Λ), see [17, Chapter I. §4.].

26

Since the homology functor sends quasi-isomorphisms to isomorphisms, itinduces a well-defined functor on D(Λ) by

Hi

Zs

f

X Y

= Hi(f)Hi(s)−1.

Lemma 0.2.15. Let P → Q → R → be a distinguished triangle in K(Λ).If P and Q are K-projective, then R is K-projective.

Proof. Let s : A → B be a quasi-isomorphism in K(Λ). Since P and Qare K-projective, ΣP and ΣQ are K-projective. This gives the followingcommutative diagram,

(ΣQ,A) //

∼=

(ΣP,A) //

∼=

(R,A) //

(Q,A) //

∼=

(P,A)

∼=

(ΣQ,B) // (ΣP,B) // (R,B) // (Q,B) // (P,B).

Therefore the map (R,A)→ (R,B) is an isomorphism by Lemma 0.1.7 andR is K-projective.

Lemma 0.2.16. Let P and Q be K-projective in K(Λ). If p : P → Q is aquasi-isomorphism, then p is an isomorphism.

Proof. The proof given here helps understanding of some of the lemmasmentioned. Since K(Λ) is triangulated, complete p : P → Q to a distin-

guished triangle Pp→ Q→ R→. Then R is K-projective by Lemma 0.2.15.

The triangle induces a long exact sequence

· · · // (R,P )∼= // (R,Q) // (R,R) // (R,ΣP )

∼= // (R,ΣQ) // · · ·

by Lemma 0.2.9. Therefore (R,R) = 0 and R ∼= 0, and p is an isomorphismby Lemma 0.2.2(iii).

Remark 0.2.17. Alternatively, by Lemma 0.2.16, the distinguished trian-

gle Pp→ Q → R → induces the long exact sequence Hi(P )

∼=→ Hi(Q) →Hi(R) → Hi−1(P )

∼=→ Hi−1(Q). Therefore Hi(R) = 0 and R is exact, andthe morphism f : 0 → R is a quasi-isomorphism by Lemma 0.1.18. SinceR is also K-projective, (R, 0) ∼= (R,R). Then there is a unique morphismg : R → 0 such that fg = 1R where 1R : R → R is the identity morphism,as shown in the following diagram.

0

f

R1R //

g??

R

27

Therefore 1R = fg = 0 as it factors through the zero object, and this givesR ∼= 0.

Definition 0.2.18. Let C be a category and S be a multiplicative systemin C. Let X be in C. A K-projective resolution of X is a K-projective Q inC, together with a morphism s : Q→ X in S.

For example, given X in K(Λ), a K-projective resolution of X is a K-projective Q in K(Λ), together with a quasi-isomorphism q : Q→ X.

Example 0.2.19. Let M be in Mod(Λ). Then the projective resolution P (M)is also a K-projective resolution. For all X in K(Λ), there is a K-projectiveresolution of X, see the (unnumbered) remark after [8, Proposition 2.12]and [43, Corollary 3.5].

Lemma 0.2.20. Let M and N be in Mod(Λ), and let p : P → M be a K-projective resolution of M . Then HomD(Λ)(M,ΣiN) ∼= HomK(Λ)(P,Σ

iN).

Proof. Let θ : HomD(Λ)(M,ΣiN)→ HomK(Λ)(P,ΣiN) and θ′ : HomK(Λ)(P,Σ

iN)→HomD(Λ)(M,ΣiN) be maps between the two Hom sets.

Given the following morphism µ =

Xs

~~

f

""M ΣiN

in HomD(Λ)(M,ΣiN), where f : X → ΣiN is any morphism in HomK(Λ)(X,ΣiN),

there is the bijective map (P, s) : (P,X)∼=→ (P,M), since s : X → M is a

quasi-isomorphism and P is K-projective. In particular, consider the mor-phism p : P → M in (P,M). Then there is a unique p′ in (P,X) such thatsp′ = p. Let θ(µ) be fp′ in HomK(Λ)(P,Σ

iN).

Conversely, given a morphism g : P → ΣiN in K(Λ), let θ′(g) be thefollowing morphism

Pp

~~

g

!!M ΣiN

in HomD(Λ)(M,ΣiN).

Finally, one can verify that θ and θ′ are well-defined and that θθ′ = θ′θ =1.

Remark 0.2.21. Alternatively, Lemma 0.2.20 can also been seen directlyfrom Lemma 0.2.12(i) and Lemma 0.2.14.

28

Let PΛ be the full subcategory of K(Λ) consisting of K-projectives, andlet S be the class of quasi-isomorphisms in K(Λ). Then S ∩ PΛ forms amultiplicative system in PΛ. Consider the following diagram,

PΛπP //

_

i

(S ∩ PΛ)−1PΛ

i

K(Λ)πC // S−1K(Λ),

where πP , πC are the canonical functors. By Lemma 0.2.12(i), quasi-isomorphismsin K(Λ) become isomorphisms in S−1K(Λ) = D(Λ), hence there is a uniquefunctor i : (S ∩ PΛ)−1PΛ → S−1K(Λ) making the diagram commutative bythe universal property of (S ∩ PΛ)−1PΛ. By Example 0.2.19, [17, Propo-sition I.3.3] and Example 0.1.8(ii), the functor i is an equivalence. ByLemma 0.2.12(ii) and Lemma 0.2.16, the functor πP : PΛ → (S ∩ PΛ)−1PΛ

is also an equivalence. Hence the functor iπP : PΛ → S−1K(Λ) is an equiv-alence. Therefore D(Λ) can be viewed either as the category of fractionsS−1K(Λ), or as PΛ, the full subcategory inside K(Λ).

By virtue of Lemma 0.1.27(i), define the functor P (−) : Mod(Λ) → K(Λ)which sends M in Mod(Λ) to the projective resolution P (M) of M in K(Λ).Similarly, define the functor I(−) : Mod(Λ) → K(Λ) which sends M inMod(Λ) to the injective resolution I(M) of M in K(Λ).

The notations P (−) and I(−) will be used for the rest of this section.

Lemma 0.2.22. Given a short exact sequence 0 → M ′µ′→ M

µ′′→ M ′′ →0 in Mod(Λ), the diagram P (M ′)

P (µ′)→ P (M)P (µ′′)→ P (M ′′)

∂→ ΣP (M ′)is a distinguished triangle in K(Λ), where the connecting morphism ∂ :P (M ′′)→ ΣP (M ′) can be described explicitly.

Proof. Consider the morphism µ′ : M ′ → M in Mod(Λ) and then the cor-responding morphism P (µ′) : P (M ′)→ P (M) in K(Λ). Complete the mor-

phism P (µ′) to a distinguished triangle P (M ′)P (µ′)→ P (M)→ C → ΣP (M ′)

in K(Λ), where C is the mapping cone E(P (µ′)) of P (µ′), since K(Λ) istriangulated.

This is portrayed in the following diagram, where the morphism ı is thecanonical inclusion and the morphism ρ is the canonical surjection.

29

P (M)

ı

= // P2//

P1//

P0//

0 //

C

ρ

= // P2 ⊕ P ′1 //

P1 ⊕ P ′0 //

P0 ⊕ 0 //

0⊕ 0 //

ΣP (M ′) = // P ′1

// P ′0// 0 // 0 //

It can be readily seen that C is a projective resolution of M ′′. Consider thefollowing diagram in K(Λ),

P (M ′)P (µ′) // P (M)

ı // Cρ //

α

ΣP (M ′)

P (M ′)P (µ′) // P (M)

P (µ) // P (M ′′)∂ // ΣP (M ′),

where the first row is the distinguished triangle constructed and the sec-ond row is the diagram described in the lemma. Since both C and P (M ′′)are projective resolutions of M ′′, by Lemma 0.1.27(i), there is a chain mapα : C → P (M ′′) which is an isomorphism by Lemma 0.2.16. Let the con-necting morphism ∂ : P (M ′′) → ΣP (M ′) be ρα−1, and since αı = P (µ)by Lemma 0.1.27(i), the diagram on the second row is then a distinguishedtriangle by (TR1).

The description of the left derived functors and right derived functors is anabridgment of [18, IV.5., 6., 7., 8.]. Let Λ1 and Λ2 be finite-dimensionalk-algebras over the field k.

Consider an additive covariant functor F : Mod(Λ1) → Mod(Λ2), and letFC : C(Λ1) → C(Λ2) be the lift of F from the module category to the

category of chain complexes, i.e. given a chain complex M = · · · → M1∂1→

M0∂0→ M−1 → · · · in C(Λ1), define the chain complex FC(M) = · · · →

FM1F∂1→ FM0

F∂0→ FM−1 → · · · in C(Λ2). Let f, g : M → N be chain mapsin C(Λ1) such that f ' g. Then it is immediate that FCf ' FCg in C(Λ2).Therefore the functor FC is well-defined on the homotopy category K(Λ1)as well. Let FK : K(Λ1) → K(Λ2) be the lift of FC from the category ofchain complexes to the homotopy category. It is also a triangulated functor.

Definition 0.2.23. In the above description, let the i-th left derived functorof F be LiF = Hi FK P .

30

K(Λ1)FK // K(Λ2)

Hi

Mod(Λ1)

P

OO

LiF //Mod(Λ2)

The following lemma views the left derived functor LiF as generalizing thefunctor F .

Lemma 0.2.24. Let us assume that F is right exact, i.e. given a short

exact sequence 0→M ′µ→M →M ′′ → 0 in Mod(Λ1), the sequence FM ′

Fµ→FM → FM ′′ → 0 is right exact in Mod(Λ2). Then

(i) LiF = 0 for i < 0,

(ii) L0F ' F ,

(iii) There is a long exact sequence

· · · // L2F (M ′′) // L1F (M ′) // L1F (M) // L1F (M ′′) //

L0F (M ′) // L0F (M) // L0F (M ′′) // 0

in Mod(Λ2).

In Lemma 0.2.24, the functor F fails to be an exact functor, and (iii) ex-presses kerFµ in terms of the long exact sequence. Consider the functori : PΛ1 → K(Λ1) and the equivalence iπP : PΛ1 → S−1K(Λ1) describedabove by taking Λ = Λ1, the functors F and FK in Definition 0.2.23and the canonical functors πi : K(Λi) → D(Λi). Let LF be the functorπ2FKi(iπP )−1 : D(Λ1)→ D(Λ2).

K(Λ1)π1 //

FK

D(Λ1)i(iπP )−1

oo

LF

K(Λ2)π2 // D(Λ2)

For example, given M in Mod(Λ), we have

HiLF (M) = HiπFK(P (M))

= HiFK(P (M))

= LiF (M).

31

In turn, the description of the right derived functors of some special functors,namely, the functors HomΛ(−, N) and HomΛ(N,−), where N is in Mod(Λ),is given.

Both the contravariant functor HomΛ(−, N) and the covariant functor HomΛ(N,−)are left exact. This allows us to give the following definition.

Definition 0.2.25. Let the i-th right derived functor RiHomΛ(−, N) ofHomΛ(−, N), written ExtiΛ(−, N), be ExtiΛ(−, N) = HiHomΛ(P (−), N) :Mod(Λ)→ Mod(Λ). Similarly, let the i-th right derived functor RiHomΛ(N,−)

of HomΛ(N,−), written ExtiΛ(N,−), be Ext

iΛ(N,−) = HiHomΛ(N, I(−)) :

Mod(Λ)→ Mod(Λ).

Lemma 0.2.26. Let A and B be in Mod(Λ). Then

(i) ([18, Proposition IV.7.2]) Given a projective module P and an injectivemodule I in Mod(Λ), we have ExtiΛ(P,B) = 0 = ExtiΛ(A, I), wherei ≥ 1,

(ii) ([18, Proposition IV.8.1]) The functors ExtiΛ(−,−) and ExtiΛ(−,−),

where i ≥ 0, are naturally equivalent.

Lemma 0.2.27. (c.f. Lemma 0.2.24) Consider a right exact sequence M ′ →M →M ′′ → 0 in Mod(Λ). Then

(i) ExtiΛ(−, N) = 0 for i < 0,

(ii) Ext0Λ(−, N) ' (−, N),

(iii) There is a long exact sequence

0 // (M ′′, N) // (M,N) // (M ′, N) //

Ext1Λ(M ′′, N) // Ext1Λ(M,N) // Ext1Λ(M ′, N) // Ext2Λ(M ′′, N) // · · ·

in Mod(Λ).

The following is consistent with Definition 0.1.24.

Definition 0.2.28. ([18, Exercise IV.8.8.]) The global dimension of Λ,denoted by gl. dim. Λ, is less than or equal to m if for all Λ-modules A,B,ExtqΛ(A,B) = 0 for all q > m. The smallest m with gl. dim. Λ ≤ m is theglobal dimension of Λ.

The following realizes the value of ExtiΛ(−, N) as the Hom space in thederived category.

32

Let M and N be in Mod(Λ).

By Lemma 0.2.6 and Lemma 0.2.20, we have

HomD(Λ)(M,ΣiN) ∼= HomK(Λ)(PM ,ΣiN),

where PM is a K-projective resolution of M,∼= H0HomΛ(PM ,Σ

iN)∼= H0ΣiHomΛ(PM , N)∼= HiHomΛ(PM , N)

≡ ExtiΛ(M,N).

Lemma 0.2.29. ([17, Chapter I. §6.]) Let 0 → X → Y → Z → 0 be ashort exact sequence of chain complexes in C(Λ). Then there is a morphism

ζ : Z → ΣX in D(Λ) such that X → Y → Zζ→ ΣX is a distinguished

triangle in D(Λ).

0.2.3 The stable category

The stable category described in this section is a different flavour of quotientcategory. For example, the left or right triangulated structure of stablecategories of an artin algebra is studied in [6]. In the interim, the followingnaıve introduction serves as a minute interlude.

The stable categories described below are attainable because given A1 andA2 in mod(Λ), the direct sum A1 ⊕ A2 is injective (resp. projective) if andonly if A1 and A2 are injective (resp. projective).

Definition 0.2.30. Given A and B in mod(Λ), let I(A,B) be the set ofhomomorphisms from A to B which factor through an injective module.

Definition 0.2.31. Given A and B in mod(Λ), let P(A,B) be the set ofhomomorphisms from A to B which factor through a projective module.

Definition 0.2.32. The (injective) stable category mod(Λ) of mod(Λ) hasthe same objects as mod(Λ), while the morphism set Hom(A,B) in mod(Λ)is defined to be Hom(A,B)/I(A,B) for all A,B in mod(Λ).

Definition 0.2.33. The (projective) stable category mod(Λ) of mod(Λ) hasthe same objects as mod(Λ), while the morphism set Hom(A,B) in mod(Λ)is defined to be Hom(A,B)/P(A,B) for all A,B in mod(Λ).

The following lemma shows that objects which are not isomorphic in mod(Λ)can be isomorphic in mod(Λ). For example, the ring Λ is isomorphic to 0 inmod(Λ).

33

Lemma 0.2.34. (c.f. Lemma 0.1.29) Let 0 be the zero object and let X bean object in mod(Λ). Then f : 0→ X is an isomorphism in mod(Λ) if andonly if X is projective.

Proof. (if ) This is immediate. (only if ) Let g : X → 0 be a morphism inmod(Λ) such that f g = 1X . This means fg − 1X is in P(X,X). Sincefg = 0, therefore 1X is in P(X,X), i.e. there is a projective module P suchthat 1X factorizes as 1X = βα as follows.

X1X //

α

X

Pβ

>>

To see that X is projective, consider an epimorphism ε : A B and amorphism θ : X → B in mod(Λ). We need to find a morphism π′ : X → Asuch that επ′ = θ.

Aε // // B

Xθ

>>

π′

``

α

P

β

OOπ

UU

Now consider the morphism θβ : P → B. Since P is projective, there is amorphism π : P → A such that επ = θβ. Therefore επα = θβα = θ1X = θ.Finally, π′ = πα is the required morphism.

0.3 Auslander-Reiten quiver

For this section, the reader is suggested to read [14] and [15] for a recapitu-lation. Let C be any category, A an abelian category and T a triangulatedcategory.

Let C′ be a full subcategory of C.

Definition 0.3.1. A morphism g : B → C in the subcategory C′ is said tobe a right almost split morphism in C′ if

34

(i) g is not a split epimorphism,

(ii) if h : C ′ → C in C′ is not a split epimorphism, then there is anh′ : C ′ → B such that gh′ = h.

The notion of a left almost split morphism in the subcategory C′ is defineddually, i.e. a morphism f : A→ B in the subcategory C′ is said to be a leftalmost split morphism in C′ if

(i) f is not a split monomorphism,

(ii) if h : A → A′ in C′ is not a split monomorphism, then there is anh′ : B → A′ such that h′f = h.

Lemma 0.3.2. ([30, Lemma 2.3]) Let the subcategory C′ be additive. Thengiven a left almost split morphism g : B → C in the subcategory C′, theendomorphism ring End(B) is a local ring.

Definition 0.3.3. A morphism h : X → Y in A is irreducible if h is neithera split monomorphism nor a split epimorphism, and if h = h2h1 then eitherh1 is a split monomorphism or h2 is a split epimorphism.

Given objectsX and Y in a Krull-Schmidt category C, the subspace rad(X,Y )of (X,Y ) is defined to be rad(X,Y ) = h ∈ (X,Y ) | 1X − gh is invertiblefor any g ∈ (Y,X). Then for m ≥ 1, the subspace radm(X,Y ) ⊆ rad(X,Y )

of rad(X,Y ) consists of all finite sums of morphisms of the form X = X0h1→

X1h2→ X2 → · · · → Xm−1

hm→ Xm = Y , where hi is in rad(Xi−1, Xi). Let dXYbe the dimension dimkIrr(X,Y ), where Irr(X,Y ) = rad(X,Y )/rad2(X,Y ).

If X and Y are indecomposable, then f : X → Y is irreducible if and onlyif f is in rad(X,Y ) but not in rad2(X,Y ). Therefore there is an irreduciblemorphism from X to Y if and only if Irr(X,Y ) 6= 0.

Definition 0.3.4. Let C be a Krull-Schmidt category. Then the Auslander-Reiten quiver of C is a quiver where the vertices are the isomorphism classes[X] of the indecomposable objects X of C, and the quiver has dXY arrowsfrom [X] to [Y ].

The following definition can be found in [14, 3.1] and in [22, Definition 1.3].

Definition 0.3.5. (c.f. Definition 2.2.2) A distinguished triangle xα→ y

β→z →, with x, y and z in the subcategory T′ of T, is an Auslander-Reitentriangle in T′ if

(i) the triangle is not split,

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(ii) if x′ is in T′, then each morphism x→ x′ which is not a split monomor-phism factors through α,

(iii) if z′ is in T′, then each morphism z′ → z which is not a split epimor-phism factors through β.

Consider the Auslander-Reiten triangle xα→ y

β→ z → in Definition 0.3.5.By Lemma 0.2.3, the morphism α is left almost split and the morphism βis right almost split.

Let us recollect some standard results with regard to Auslander-Reiten tri-angles.

Lemma 0.3.6. ([30, Lemma 2.6]) Let xα→ y

β→ z → be a distinguishedtriangle in T. Suppose β is right almost split. Then the following are equiv-alent.

(i) End(x) is local,

(ii) β is right minimal,

(iii) α is left almost split,

(iv) The distinguished triangle is an Auslander-Reiten triangle.

The following lemma shows how the Auslander-Reiten triangles can be in-herent in the Auslander-Reiten quiver.

Lemma 0.3.7. Let T be Krull-Schmidt. If x→ y → z → is an Auslander-Reiten triangle in T, then given an indecomposable object yi in T, the fol-lowing are equivalent.

(i) There is an irreducible morphism x→ yi,

(ii) There is an irreducible morphism yi → z,

(iii) yi is an indecomposable direct summand of y.

Proof. This is described in [15, 4.8].

Definition 0.3.8. [39, I.1.] Let A be k-linear and Hom finite. A right Serrefunctor is an additive functor S : A→ A, together with isomorphisms

ϕA,B : (A,B)∼=→ (B,SA)∗

for any A,B ∈ A, which are natural in A and in B, and where (−)∗ =Homk(−, k). If S is an autoequivalence, then it is a Serre functor.

36

Let ϕA,A(idA) be denoted by ϕA. Any two right Serre functors are isomor-phic, and if ε is an autoequivalence of A, then εS ∼= Sε, see [32, Section3].

Definition 0.3.9. The triangulated category T is said to have right (resp.left) Auslander-Reiten triangles if given any indecomposable z (resp. x) inT, there is an Auslander-Reiten triangle x → y → z → Σx. It is then saidto have Auslander-Reiten triangles if it has both right and left Auslander-Reiten triangles.

Lemma 0.3.10. The following are equivalent.

(i) T has a Serre functor,

(ii) T has Auslander-Reiten triangles.

Proof. This is [39, Theorem I.2.4]. In addition, the Auslander-Reiten trian-gle is of the form SΣ−1z → y → z → Sz, where S is the Serre functor andΣ is the translation functor of T.

This leads us to the following definition.

Definition 0.3.11. The functor τ ∼= SΣ−1 is the Auslander-Reiten trans-lation.

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Chapter 1

The approximationproperties of subcategories

1.1 Introduction

The idea of approximation is very frequent in mathematics indeed. Severalnotions in Chapter 0 are inspired in such a way. For example, given A inMod(Λ), a projective resolution P (A) of A intuitively approximates howprojective A is. The right derived functor ExtiΛ(−, N) of HomΛ(−, N) inSection 0.2.2 can also be understood to approximate the notion of a cok-ernel. The projective stable category of a given category in Section 0.2.3approximates the original category by identifying the projective objects withthe zero objects.

Paradoxical as it seems, the greater the degree of accuracy to be attained,the greater the strength and the more varieties of approximation needed. Inthis chapter, torsion pairs in abelian categories and torsion theories in trian-gulated categories are introduced. The existence of certain adjoint functorsin triangulated categories is also studied. Intuitively, they are all differentexpressions of subcategories approximating their ambient categories. Thechapter goes on to introduce some examples of torsion theories, namely t-structures and split torsion theories, and finishes with a characterization ofa split torsion theory and a classification of split torsion theories in a chosenderived category.

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1.2 Existence of certain adjoint functors

Let Λ be a finite-dimensional k-algebra over the field k. As usual, let Mod(Λ)be the category of Λ-left-modules. In this section, let T be a triangulatedcategory with translation functor Σ, and X and Y be full additive subcat-egories of T . The subcategories X and Y are assumed to be closed underisomorphisms.

Let us begin with torsion pairs in abelian categories.

Definition 1.2.1. [1, Definition VI.1.1] A pair (M,N ) of full subcategoriesof Mod(Λ) is a torsion pair if

(i) Hom(M,N) = 0 for all M ∈M, N ∈ N ,

(ii) Hom(M,−)|N = 0 implies M ∈M,

(iii) Hom(−, N)|M = 0 implies N ∈ N .

Here, M is the torsion class and N is the torsion-free class. By virtue of(ii) and (iii), M and N uniquely determine each other.

Definition 1.2.2. [1, Proposition VI.1.4] Let U be in Mod(Λ) and tU bethe trace ofM in U , that is, the sum of images of all homomorphisms frommodules in M to U . Since M is closed under images and direct sums, tUis the largest submodule of U that lies in M.

Remark 1.2.3. ([1, Propositions VI.1.4, 1.5]) Intuitively, the functor t inDefinition 1.2.2 is an indicator which measures how far U is from being inM or in N . It has the following significance.

(i) (approximation) It approximates U to give tU in M,

(ii) (characterization) It characterizes the torsion and torsion-free classes,sinceM = M ∈ Mod(Λ) | tM = M and N = N ∈ Mod(Λ) | tN =0,

(iii) (existence and uniqueness) It gives existence and uniqueness of shortexact sequences of the form 0→ tU → U → U

tU → 0.

Instead of the decomposition of an object into direct summands, the fol-lowing description of torsion theories in triangulated categories intuitivelydecomposes a category into a pair of subcategories. The definition givenbelow is the one in [21, Definition 2.2], except that we do not assume X andY to be closed under direct sums and direct summands.

39

Definition 1.2.4. Let X⊥ = t ∈ T |(x, t) = 0 for all x in X and ⊥X =t ∈ T |(t, x) = 0 for all x in X.

Definition 1.2.5. (X , Y) forms a torsion theory if

(i) (X , Y) = 0,

(ii) T = X ∗ Y, i.e. given t in T , there is a torsion theory triangle x →t→ y → Σx with x in X and y in Y.

Remark 1.2.6. (i) In Definition 1.2.5, (i) gives X ∩ Y = 0. Otherwise anon-zero U in X ∩ Y would give idU in (X , Y), which is not possible.Accordingly, there are no non-zero U in X and V in Y such thatU ∼= V .

(ii) Let xµ→ t

ν→ y → Σx be the torsion theory triangle described in (ii).Then by Lemma 0.2.2(i), the composition νµ is zero, which accordswith (i). By Lemma 0.2.2(iv), given t ∼= x⊕ y with t, x, y in T , X , Yrespectively, there is always a distinguished triangle x→ t→ y → Σx,thus satisfying (ii) for the chosen t.

The following lemma gives the situations where X and Y uniquely determineeach other.

Lemma 1.2.7. Let (X , Y) be a torsion theory.

(i) If ΣX ⊆ X , then X⊥ = Y and ⊥Y = X .

(ii) If X and Y are closed under direct summands, then X⊥ = Y and⊥Y = X .

Proof. (i) This is [7, Remark I.2.2].

(ii) It is immediate that Y ⊆ X⊥. To see that X⊥ ⊆ Y, consider t inX⊥. Since (X , Y) is a torsion theory, there is a distinguished trianglex → t → y → Σx with x in X and y in Y, where the morphismx → t is zero. Then by Lemma 0.2.2(v), the distinguished trianglet → y → Σx → Σt splits, i.e. y ∼= t ⊕ Σx. Since Y is closed underdirect summands and isomorphisms, t is in Y. Hence X⊥ ⊆ Y and soX⊥ = Y. Similarly, ⊥Y = X .

Torsion theories in triangulated categories are analogues of torsion pairs inabelian categories. Given a torsion pair (M, N ), the functor t described inDefinition 1.2.2 characterizes the subcategoriesM and N , see Remark 1.2.3.

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The functor t involves the notion of images, which does not necessarily makesense in triangulated categories. Following on from this, an analogue of thefunctor t in torsion theories is not immediate, and this leads to the study ofcertain adjoint functors in triangulated categories.

Definition 1.2.8. An X -precover for an object t in T is a morphism α :x → t for some x in X , such that for all x′ in X , each morphism x′ → tfactorizes through α. An X -cover is an X -precover which is right minimal.The notion of an X -(pre)envelope is defined dually.

Definition 1.2.9. X is said to be a (pre)covering for T if every object inT has an X -(pre)cover. The notion of a (pre)enveloping for T is defineddually.

Example 1.2.10. (i) In Remark 1.2.3(i), tU → U is an M-cover.

(ii) Let (X , Y) be a torsion theory. Then X is a precovering and Y is apreenveloping for T . This is because given t in T , there is a distin-guished triangle x→ t→ y → Σx with x in X , y in Y, where x→ t isan X -precover and t→ y is a Y-preenvelope (Lemma 0.2.2(vi)).

The notion of adjoints is given in Example 0.1.17(iv).

Lemma 1.2.11. Let L be the embedding functor ı : X → T . Assume theright adjoint R : T → X exists. Let x be in X and t be in T . Consider the

isomorphism τ : (x, t)∼=→ (x,Rt) and the morphism εt = τ−1(idRt). Then

given f : x → t, there is a unique morphism f ′ : x → Rt such that thefollowing diagram commutes.

x

f

f ′

~~Rt

εt // t

In addition to this, the morphism εt = τ−1(idRt) is right minimal.

Proof. By the naturality of τ , the morphism f ′ = τ(f) is the required uniquemorphism, and this is also conceived of as the universal property of the counitεt : Rt→ t. The morphism εt is also right minimal. Indeed suppose εtg = εtfor any given g : Rt→ Rt and consider the following commutative diagramobtained from the naturality of τ ,

(Rt,Rt)g∗ //

τ−1

(Rt,Rt)

τ−1

(Rt, t)

g∗ // (Rt, t).

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Therefore (τ−1(idRt))g = τ−1(idRtg), which gives εtg = τ−1(g) and thenτ−1(g) = τ−1(idRt). Since τ−1 is an isomorphism, g = idRt which is anautomorphism. Hence εt is right minimal.

Remark 1.2.12. In Lemma 1.2.11, X is covering in T . Intuitively, Rt in Xis the approximation of t in X , i.e. the right adjoint R can be understoodas measuring how close X is in approximating T .

Definition 1.2.13. ([5]) Let (X , Y) be a torsion theory. If ΣX ⊆ X , then(X , Y) is a t-structure.

Remark 1.2.14. Let (X , Y) be a t-structure. Then by Lemma 1.2.7(i), Xand Y uniquely determine each other.

Lemma 1.2.15. Let (X , Y) be a t-structure. Then Σ−1Y ⊆ Y.

Proof. This is [7, Remark I.2.2].

Lemma 1.2.16. Let (X , Y) be a t-structure. Then for each t in T , thereis a distinguished triangle x → t → y → Σx, with x in X and y in Y, andthe third morphism y → Σx is uniquely determined.

Proof. Suppose there are distinguished triangles xα→ t

β→ yγ1→ Σx and

xα→ t

β→ yγ2→ Σx. Then consider the following diagram,

xα // t

β // yγ1 //

g

Σx

xα // t

β // yγ2 // Σx.

By (TR3), there is a morphism g : y → y such that γ1 = γ2g and gβ = β.Let f = idy − g : y → y. Then fβ = (idy − g)β = β − gβ = 0.

Now consider the following diagram,

tβ //

yγ1 //

f

Σx−Σα //

h

Σt

0 // y

id // y // 0.

Since fβ = 0, by (TR3) there is h : Σx → y such that hγ1 = idyf = f .Since (Σx, y) = 0, h = 0, and so f = idy − g = 0, g = idy and γ1 = γ2.

Lemma 1.2.17. Let U = X ∗ Y. If ΣX ⊆ X and ΣY ⊆ Y, then ΣU ⊆ U .

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Proof. This is immediate.

Definition 1.2.18. A subcategory X is closed under extensions if given tin T and a distinguished triangle x1 → t→ x2 → Σx1 with x1 and x2 in X ,then t is in X .

Lemma 1.2.19. The subcategories X⊥ and ⊥Y are closed under extensions.

Proof. Let u be in Y. Given t in T , consider a distinguished triangle x1 →t → x2 → Σx1 with x1 and x2 in ⊥Y. Then by Lemma 0.2.9, the sequence(x2, u)→ (t, u)→ (x1, u) is exact. Since (x2, u) = (x1, u) = 0, it follows that(t, u) = 0 and so t is in ⊥Y and ⊥Y is closed under extensions. Similarly,X⊥ is closed under extensions.

Example 1.2.20. Let (X , Y) be a t-structure. By Remark 1.2.14, X = ⊥Ywhich, by Lemma 1.2.19, is closed under extensions. For example, given tin T and a distinguished triangle x1 → x2 → t→ Σx1 with x1 and x2 in X ,then t is in X .

Remark 1.2.21. (i) Let X be closed under extensions. Given x in X andthe isomorphism x′ ∼= x, embed the isomorphism into a distinguished

triangle x∼=→ x′ → z → Σx in T . Then by Lemma 0.2.2(iii), z ∼= 0.

Since X is closed under isomorphisms, z is in X . Since both x andz are in X and X is closed under extensions, x′ is in X . The twoconditions, that X is closed under extensions, and that X is closedunder isomorphisms, are compatible.

(ii) If X is closed under extensions, then X is closed under direct sums. Letx1 and x2 be in X . Then by Lemma 0.2.2(iv), there is a distinguishedtriangle x1 → x1⊕x2 → x2 → Σx1. Since X is closed under extensions,x1 ⊕ x2 is in X .

Definition 1.2.22. Let u and v be in T . A morphism f : u→ v is an X -leftphantom if given any x in X and any morphism g : x→ u, the compositionfg is zero. Dually, a morphism f : u→ v is a Y-right phantom if given anyy in Y and any morphism g : v → y, the composition gf is zero.

For the rest of the chapter, given a torsion theory (X , Y), assume X and Yto determine each other uniquely. By Lemma 1.2.19, the subcategories Xand Y are both closed under extensions and direct sums (Remark 1.2.21(ii)).The following propositions give necessary and sufficient conditions for theexistence of certain adjoint functors in triangulated categories.

Proposition 1.2.23. (c.f. [27, 1.1], [7, Proposition I.2.3]) Let (X , Y) bea torsion theory. Then the inclusion ı : X → T admits a right adjoint R: T → X if and only if for each t in T , there is a distinguished triangle

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x→ t→ y → Σx, where x ∈ X and y ∈ Y, and the morphism h: Σ−1y → xis an X -left phantom.

x′

0

!!Σ−1y // x

Proof. (if ) Let t be in T . Then there is a distinguished triangle x → t →y → Σx, with x ∈ X and y ∈ Y, and the morphism h : Σ−1y → x is anX -left phantom. Let x′ be in X . Applying the homological functor (x′,−)to the distinguished triangle, there is the long exact sequence (x′,Σ−1y)→(x′, x) → (x′, t) → (x′, y) with (x′, y) = 0 (Lemma 0.2.9). The first mapis zero since the morphism h: Σ−1y → x is an X -left phantom. Hence(x′, x) ∼= (x′, t) and R exists by defining R(t) = x. (only if ) Since the right

adjoint R exists, there is the isomorphism τ−1 : (x,Rt)∼=→ (x, t) for all x

in X , t in T . Let α = τ−1(idRt) : Rt → t. Then by Lemma 1.2.11, α isright minimal. Embed α into a distinguished triangle Rt

α→ t → y → ΣRt.By [22, Lemma 2.1], y is in Y. Let x′ be in X . Applying the homologicalfunctor (x′,−) to the distinguished triangle, there is the exact sequence(x′,Σ−1y)→ (x′, Rt)→ (x′, t) (Lemma 0.2.9). By Lemma 1.2.11, the secondmap is an isomorphism, and so the first map is zero. Hence the morphismh: Σ−1y → Rt is an X -left phantom.

The following is the dual.

Proposition 1.2.24. Let (X , Y) be a torsion theory. Then the inclusionı : Y → T admits a left adjoint L : T → Y if and only if for each t in T ,there is a distinguished triangle x→ t→ y → Σx, where x ∈ X and y ∈ Y,and the morphism h: y → Σx is a Y-right phantom.

y //

0

Σx

y′

Proof. This is the dual of Proposition 1.2.23.

Example 1.2.25. Let (X , Y) be a t-structure. Then the inclusion ı : X → Tadmits a right adjoint R : T → X . Similarly, the inclusion ı : Y → Tadmits a left adjoint L : T → Y.

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Proof. Given t in T , there is a distinguished triangle x → t → y → Σx,where x ∈ X and y ∈ Y. By Lemma 1.2.15, Σ−1y is in Y and so (x′,Σ−1y) =0 for any x′ in X and h : Σ−1y → x is trivially an X -left phantom. Thereforeby Proposition 1.2.23, the inclusion ı : X → T admits a right adjoint R :T → X . Similarly, by Proposition 1.2.24, the inclusion ı : Y → T admits aleft adjoint L : T → Y.

Remark 1.2.26. In Example 1.2.25, given t in T , there is a unique torsiontheory triangle of the form Rt → t → Lt → ΣRt. This makes sense,since any two right (resp. left) adjoints of the same functor are naturallyequivalent.

Corollary 1.2.27. Let (X , Y) be a torsion theory. Suppose the right adjointR in Proposition 1.2.23 and the left adjoint L in Proposition 1.2.24 exist.Then X = t ∈ T | Lt = 0 and Y = t ∈ T | Rt = 0.

Proof. Let x be in X and t be in T . If Rt = 0, then t ∈ X⊥ = Y, since(ıx, t) ∼= (x,Rt) = (x, 0) = 0. On the other hand, given t ∈ Y = X⊥, since0 = (ıx, t) ∼= (x,Rt) for each x in X , therefore Rt = 0. Hence Y = t ∈ T |Rt = 0. Similarly, X = t ∈ T | Lt = 0.

By virtue of Example 1.2.25, Remark 1.2.26 and Corollary 1.2.27 are sum-marized in the following.

Property 1.2.28. (c.f. Remark 1.2.3) Let (X , Y) be a t-structure andconsider the right adjoint R and the left adjoint L in Example 1.2.25. Then

(i) (existence and uniqueness) For each t in T , there is a torsion the-ory triangle Rt → t → Lt → ΣRt, and it is unique up to a uniqueisomorphism,

(ii) (characterization) X = t ∈ T | Lt = 0 and Y = t ∈ T | Rt = 0.

Example 1.2.29. Let (X , Y) be a t-structure with X , Y closed under directsummands. Let t be in X⊥. Then since (x, 0) ∼= (x, t) = 0, thereforeRt = 0. This gives the distinguished triangle 0 → t → Lt → 0. ByLemma 0.2.2(iii), t ∼= Lt, and the above distinguished triangle is isomorphic

to tid→ t → 0 → Σt. Alternatively, let x → t → y → Σx be the torsion

theory triangle, where x is in X and y is in Y. Since t is in X⊥, thereforey ∼= t ⊕ Σx by Lemma 0.2.2(v). It follows that Σx is in Y. On the otherhand, since (X , Y) is a t-structure, Σx is in X as well, therefore Σx is inX ∩ Y = 0. This gives Σx ∼= 0 and t ∼= y. Then the torsion theory triangleis isomorphic to the distinguished triangle 0 → t

α→ t → 0 where α is anisomorphism.

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Remark 1.2.30. (c.f. Remark 1.2.3, Remark 1.2.12) Intuitively, a torsiontheory (X , Y) can be understood as measuring how far each t in T is frombeing in X or in Y. Since (X , Y) is a torsion theory, X ∩Y = 0 and so thereis no contradiction (Remark 1.2.6(i)).

1.3 Examples of torsion theories

In this section, let k be a field and T be a k-linear Hom finite and Krull-Schmidt triangulated category with translation functor Σ and Serre functorS, see Definition 0.3.8. Let X ,Y,Z be full additive subcategories of T . Theyare assumed to be closed under isomorphisms.

Let us begin with a few lemmas.

Lemma 1.3.1. (i) If S−1ΣX ⊆ X , X ⊆ S−1X , then ΣX ⊆ X .

(ii) If S−1ΣX ⊆ X , X ⊆ ΣX , then S−1X ⊆ X .

(iii) If ΣX ⊆ X , S−1X ⊆ X , then S−1ΣX ⊆ X .

(iv) If Y ⊆ S−1ΣY, ΣX ⊆ Y, then SX ⊆ Y.

Proof. (i) This is because ΣX ⊆ ΣS−1X = S−1ΣX ⊆ X .

(ii) This is because S−1X ⊆ Σ−1X ⊆ X .

(iii) This is because S−1ΣX ⊆ S−1X ⊆ X .

(iv) Since ΣX ⊆ Y ⊆ S−1ΣY, therefore X ⊆ S−1Y, SX ⊆ Y.

Lemma 1.3.2. Suppose X⊥ = Y, X = ⊥Y and (X ,Y) = 0. Then SY ⊆ Yif and only if S−1X ⊆ X .

Proof. (only if) Let X be in X and Y be in Y. Since SY ⊆ Y, thereforeSY = Y ′ for some Y ′ in Y. Hence

(S−1X,Y ) ∼= (X,SY )

= (X,Y ′)

= 0.

Therefore S−1X ∈ ⊥Y = X . (if) Similar.

Lemma 1.3.3. Suppose X⊥ = Y, X = ⊥Y and (X ,Y) = 0. Then SX ⊆ Xif and only if S−1Y ⊆ Y.

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Proof. Similar to Lemma 1.3.2.

Consider t1 and t2 in T with t1 ∼= x1⊕y1 and t2 ∼= x2⊕y2⊕z, where x1 andx2 are in X , y1 and y2 are in Y and z is in Z. In both cases objects fromboth X and Y are present. However, there might or might not be the twosubcategories U1 and U2 of T such that each t in T can be written t ∼= u1⊕u2

with u1 in U1 and u2 in U2. A split torsion theory is an expression for this.

Definition 1.3.4. A split torsion theory is a torsion theory where all thetorsion theory triangles split.

Theorem 1.3.5. (Characterizations of split torsion theories) Let (X ,Y) bea torsion theory. Then the following statements are equivalent.

(i) (X ,Y) is a split torsion theory,

(ii) (Y,ΣX) = 0 for all X ∈ X , Y ∈ Y,

(iii) S−1ΣX ⊆ X ,

(iv) SΣ−1Y ⊆ Y.

Proof. Let x ∈ X and y ∈ Y.

(i) ⇒ (ii): Consider f : y → Σx. Complete it to a distinguished triangle

x → t → yf→ Σx in T . Then f = 0 by definition of a split torsion theory

(Lemma 0.2.2(v)).

(ii) ⇒ (i): This is true by definition (Lemma 0.2.2(v)).

(ii) ⇒ (iii): Since (S−1Σx, y) ∼= (y,Σx)∗ = 0, therefore S−1Σx ∈ ⊥Y = X .

(iii)⇒ (ii): For x′ = S−1Σx ∈ X , we have (y,Σx) ∼= (Σx, Sy)∗ ∼= (S−1Σx, y)∗ =(x′, y)∗ = 0.

(ii) ⇒ (iv): Since (x, SΣ−1y) ∼= (Σ−1y, x)∗ ∼= (y,Σx)∗ = 0, thereforeSΣ−1y ∈ X⊥ = Y.

(iv)⇒ (ii): For y′ = SΣ−1y ∈ Y, we have (y,Σx) ∼= (Σ−1y, x) ∼= (x, SΣ−1y)∗ =(x, y′)∗ = 0.

Remark 1.3.6. The composition SΣ−1 is the Auslander-Reiten translationgiven in Definition 0.3.11.

Example 1.3.7. (c.f. Example 1.2.25) Let (X , Y) be a split torsion theory.Then the inclusion ı : X → T admits a right adjoint R : T → X . Similarly,the inclusion ı : Y → T admits a left adjoint L : T → Y.

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Proof. Consider the morphism h : Σ−1y → x in Proposition 1.2.23. Since(Σ−1y, x) = (y,Σx) = 0 by Theorem 1.3.5, h is trivially an X -left phantom.Therefore by Proposition 1.2.23, the inclusion ı : X → T admits a rightadjoint R : T → X . Similarly, by Proposition 1.2.24, the inclusion ı :Y → T admits a left adjoint L : T → Y.

Remark 1.3.8. By virtue of Example 1.3.7, Property 1.2.28 is also true forsplit torsion theories.

Remark 1.3.9. Consider the diagram in Proposition 1.2.23. If the torsiontheory (X , Y) is a t-structure, then the morphism x′ → Σ−1y is zero andthe right adjoint exists (Example 1.2.25). On the other hand, Theorem 1.3.5describes the behaviour of the morphism Σ−1y → x, and characterizes an-other example of a torsion theory, the split torsion theory, where the rightadjoint exists as well (Example 1.3.7).

Lemma 1.3.10. Let (X ,Y) be a split torsion theory. Then for each inde-composable t in T , either t is in X or t is in Y.

Proof. Since (X ,Y) is a torsion theory, X ∩ Y = 0 and so t cannot be inboth X and Y (Remark 1.2.6(i)). Given an indecomposable t in T , there isa distinguished triangle x→ t→ y → Σx for some x in X and y in Y. Sincethe torsion theory is split, t ∼= x⊕y. Since t is indecomposable, either x ∼= 0or y ∼= 0, which gives either t ∼= y or t ∼= x. Therefore either t is in X or t isin Y, since X and Y are closed under isomorphisms.

Remark 1.3.11. The converse of Lemma 1.3.10 is not true. Suppose (X ,Y)is a torsion theory, and that for each indecomposable t in T , either t is inX or t is in Y. Then (X ,Y) does not need to be a split torsion theory. Thisis because the existence of a torsion theory triangle is not unique.

Lemma 1.3.12. Let (X ,Y) be a split torsion theory. Then (X ,Y) is at-structure.

Proof. Consider an indecomposable x in X . By Lemma 1.3.10, either Σx isin X or Σx is in Y. Suppose Σx is in Y. By Theorem 1.3.5, Y ⊆ S−1ΣY,which gives Sx ∈ Y (Lemma 1.3.1(iv)). By Lemma 0.3.10, there is theAuslander-Reiten triangle u → v → x → Sx in T . However, the trianglewould split, since x is in X and Sx is in Y, which is not possible. ThereforeΣx is in X , i.e. ΣX ⊆ X .

1.4 The finite derived category of Dynkin type

Let 4 be a quiver (a directed graph). Given an arrow α, the initial point(resp. end point) of α is denoted by s(α) (resp. e(α)). A path of length

48

l ≥ 1 from a vertex x to a vertex y is of the form (x | α1, . . . , αl | y) witharrows αi satisfying e(αi) = s(αi+1) for 1 ≤ i < l, s(α1) = x and e(αl) = y.For any vertex x in 4, a path of length 0 is denoted by (x | x). Let the pathalgebra k4 be the k-vector space with basis the set of all paths of lengthl ≥ 0 in 4. The product of two paths p1 = (x1 | α1, . . . , αm | y1) andp2 = (x2 | β1, . . . , βn | y2) is p1p2 = (x1 | α1, . . . , αm, β1, . . . , βn | y2) ify1 = x2. Otherwise it is zero. The global dimension of a path algebra k4is either 0 or 1 (Example 0.1.25).

Below are the cases where 4 is a Dynkin graph.

The Dynkin graph of type An:

1 2 3 · · · n− 1 n

The Dynkin graph of type Dn, n ≥ 4:

n− 1

1 2 3 · · · n− 2

n

The Dynkin graph of type E6:

6

1 2 3 4 5


7

1 2 3 4 5 6


8

1 2 3 4 5 6 7

49

The finite derived category Db(mod k4) is a k-linear Hom finite triangulatedcategory. It is Krull-Schmidt as well. By [14, Theorem 3.6], the finite derivedcategory Db(mod k4) has Auslander-Reiten triangles, and by Lemma 0.3.10also a Serre functor.

Given a quiver 4, there is the translation quiver Z4 naturally induced by4. The vertices of Z4 is the set Z×40, where 40 is the set of vertices of4. The number of arrows from (n, x) to (m, y) in Z×40 is the number ofarrows from x to y (resp. from y to x) in 40 if n = m (resp. m = n + 1),and there are no arrows otherwise. If 4 is a Dynkin graph, then Z4 doesnot depend on the orientation of4, see [15, I.5.6], and the Auslander-Reitenquiver of the finite derived category Db(mod k4) is Z4. This is given in[14, 4.5].

Lemma 1.4.1. Let x and y be indecomposable objects of the finite derivedcategory Db(mod k4), where 4 is a Dynkin graph. Then by [14, 4.6], anynon-zero morphism f from x to y is a linear combination of morphisms,written f = Σαifi, where the αi are scalars and the fi : x→ y are composi-tions of irreducible morphisms.

Let 4 be a Dynkin graph. Then the category indDb(mod k4) is equivalentto the mesh category of Z4. This means that given an Auslander-Reiten

triangle aα→ b

β→ c→ in the finite derived category Db(mod k4), which canbe read off from the Auslander-Reiten quiver, there is the relation βα = 0(Lemma 0.2.2(i)), and only relations of this kind are present. The aboveis only a gentle and informal initiation, and the reader will be able to seekguidance from [14, 4.6].

Together with Lemma 1.4.1, the finite derived category Db(mod k4), where4 is a Dynkin graph, is then said to be standard.

For example, the Auslander-Reiten quiver of Db(mod kAn) is

•

•

n

•

•

•

??

•

??

•

??

•

??

•

??

. . . •

??

•

??

•

??

•

??

• . . .

•

??

•

??

•a

??

•

??

•

??

2

??

•

??

•

??

•

??

•

1

??

•

??

•

??

•

??

•

??

The following coordinate system on the quiver is employed. Suppose the

50

indecomposable object a has coordinates (i, j). The coordinates of some ofits surrounding objects are given in the following diagram.

(i− 2, j − 1)

(i− 1, j)

(i, j + 1)

(i− 2, j − 2)

??

(i− 1, j − 1)

??

(i, j)

??

(i+ 1, j + 1)

(i− 1, j − 2)

??

(i, j − 1)

??

(i+ 1, j)

??

The coordinates on the bottom line of the quiver satisfy the equation y−x =2, and those on the top line satisfy y − x = n+ 1.

Let Σ be the translation functor of Db(mod kAn). Then the action of Σ isgiven by Σ(i, j) = (j − 1, i + n + 2) ([25, Example 2.8]), and the action ofthe Auslander-Reiten translation τ by τ(i, j) = (i− 1, j − 1).

Definition 1.4.2. Let a be an indecomposable object of Db(mod kAn), andlet L(a) be the set of indecomposable objects with non-zero morphisms toa. Dually, let R(a) be the set of indecomposable objects to which there arenon-zero morphisms from a.

Sketch 1.4.3. The two regions L(a) and R(a) are depicted below. Thecoordinates of the corners of R(a) are given by a = (i, j), b = (j − 2, j),c = (j − 2, i+ n+ 1) and d = (i, i+ n+ 1).

•d′ •dy−x=n+1

•c′ •c

L(a) •a R(a)

•b′ •by−x=2

Example 1.4.4. Consider the Auslander-Reiten quiver of Db(mod kA4) andthe indecomposable object c2.

51

•m4

•a4

α′3 •b4

β′3

•c4

•d4

. . . •m3

??

•a3

α′2

α3??

•b3

β′2

β3??

•c3

γ3??

•d3

??

• . . .

•a2

α′1

α2??

•b2

β′1

β2??

•c2

γ2??

•d2

??

•

??

•a1

α1??

•b1β1??

•c1γ1??

•d1

??

•

??

•

Since b2 → b3⊕c1 → c2 → is an Auslander-Reiten triangle, β′2β2 +γ1β′1 = 0,

and so by Lemma 1.4.1, (b2, c2) is one-dimensional. In general, given inde-composable objects x and y of Db(mod kA4), if (x, y) is non-zero, then (x, y)is one-dimensional. Since b1 → b2 → c1 → is an Auslander-Reiten triangle,β′1β1 = 0. Therefore by Lemma 1.4.1, (b1, c1) = 0 and so (b1, c2) = 0,and then (a2, c2) = (m3, c2) = 0. Similarly, since m4 → a3 → a4 →is an Auslander-Reiten triangle, by Lemma 1.4.1, (m4, a4) = 0 and so(m4, c2) = 0. Accordingly, L(c2) = a3, a4, b2, b3, c1, c2. Dually, R(c2)= c2, c3, c4, d1, d2, d3.

For the rest of this section, let X ,Y be full additive subcategories of thefinite derived category Db(mod kAn), denoted by T , with translation func-tor Σ, Serre functor S and Auslander-Reiten translation τ ∼= SΣ−1. Thesubcategories are assumed to be closed under isomorphisms and direct sum-mands.

Definition 1.4.5. A zig zag Z in the Auslander-Reiten quiver of the finitederived categoryDb(mod kAn) is a set of coordinates (x1, y1), (x2, y2), . . . , (xn, yn)where y1−x1 = 2, yn−xn = n+1 and (xi+1−xi, yi+1−yi) ∈ (−1, 0), (0, 1)for 1 ≤ i ≤ n− 1.

Proposition 1.4.6. (c.f. Lemma 1.3.10) Let (X ,Y) be a split torsiontheory. If the indecomposable u = (g, h) is in X , then the region Ru =(x, y)|g ≤ x, h ≤ y and 2 ≤ y − x ≤ n+ 1 is in X .

u Ru

52

Proof. Since (X ,Y) is a torsion theory, X is closed under extensions. In thefollowing diagram, let v1 = (g, h+ 1), v2 = (g+ 1, h) and w = (g+ 1, h+ 1).Then there is the Auslander-Reiten triangle u → v1 ⊕ v2 → w →. Sinceτ−1X ⊆ X by Theorem 1.3.5, w = τ−1(u) is in X . Since X is closedunder extensions and direct summands, v1 and v2 are in X . By repeating a(similar) argument on v1 and v2 and so on, the two line segments t1 and t2are in X . Eventually, the region Ru is in X , since τ−1X ⊆ X .

t1

v1

u w Ru

v2

t2

Corollary 1.4.7. Let (X ,Y) be a split torsion theory. If X is neither zeronor all of T , then there is a zig zag Z such that indX = τ−iZ for i ≥ 0 andindY = τ jZ for j > 0. One example is as follows.

Y X

Proof. Consider a horizontal line y−x = k with 3 ≤ k ≤ n in the Auslander-Reiten quiver of T . If there are objects from X on this line, then there isa leftmost such object. Otherwise there are objects of X arbitrarily farto the left on y − x = k, so all objects on y − x = k are in X becauseτ−1X ⊆ X by Theorem 1.3.5. In the following diagram, let d1 = (u1, u2)and d2 = (u1+1, u2−1) be objects on the lines y−x = k+1 and y−x = k−1respectively. Then there is the Auslander-Reiten triangle d0 → d1 ⊕ d2 →d′0 →, where d0 = (u1, u2 − 1) and d′0 = (u1 + 1, u2). Since d0 and d′0both lie on the line y − x = k which is in X , it follows that d1 and d2 arein X , since X is closed under extensions and direct summands. Thereforethe two neighbouring lines y − x = k + 1 and y − x = k − 1 are in X .

53

Repeating the argument for other (horizontal) lines, X has to contain allthe indecomposable objects of T , i.e. X has to be all of T . The cases wherek = 2 and k = n+ 1 are similar.

d1

d0

??

d′0

d2

??

Now on the line y−x = k, 2 ≤ k ≤ n+1, let ak be the leftmost object whichis in X , as shown in the following diagram. Suppose that for some k theobject ak+1 was neither a+

k nor a−k . If ak+1 were left of a−k , then its regionRak+1

⊆ X (Proposition 1.4.6) would contain an object to the left of ak ony − x = k, which is a contradiction. Also ak+1 could not have been to theright of a+

k since a+k is in the region Rak ⊆ X (Proposition 1.4.6). Therefore

ak+1 can only be a+k or a−k .

a−k

a+k

ak

??

Finally, the result follows by applying Proposition 1.4.6 on all such leftmostobjects.

Theorem 1.4.8. (classification of split torsion theories) There is a bijectionbetween the set of zig zag Z’s and the set of split torsion theories (X ,Y) inthe finite derived category Db(mod kAn), where X and Y are separated by Zsuch that indX = τ−iZ for i ≥ 0 and indY = τ jZ for j > 0.

Proof. If (X ,Y) is a split torsion theory, then by Corollary 1.4.7, thereis a zig zag Z which separates X and Y on the Auslander-Reiten quiver.Conversely, suppose there is a zig zag Z which separates X and Y on the

54

Auslander-Reiten quiver. Then by inspection, Y (resp. X ) is precisely X⊥(resp. ⊥Y) (Sketch 1.4.3). Therefore by Lemma 1.2.19, X is closed underextensions. The subcategory X is trivially precovering since (X , Y) = 0.Therefore (X ,Y) is a torsion theory by [21, Proposition 2.3]. Since it canbe readily seen by considering the Auslander-Reiten quiver that τ−1X ⊆ X ,by Theorem 1.3.5, (X ,Y) is a split torsion theory.

The following lemma is a special case of Lemma 1.3.12. This is because thefinite derived category Db(mod kAn) has a Serre functor.

Lemma 1.4.9. Let (X ,Y) be a split torsion theory. Then (X ,Y) is a t-structure.

Proof. Let u = (g, h) be in X . Then by Proposition 1.4.6, the region Ru isin X . Thus Σu = Σ(g, h) = (h− 1, g+n+ 2) is in Ru = (x, y)|g ≤ x, h ≤ yand 2 ≤ y − x ≤ n+ 1 ⊆ X .

An alternative proof is possible by inspecting the action of the Serre functorS and by appealing to Lemma 1.3.1(i). First we need a little lemma.

Lemma 1.4.10. Let (X ,Y) be a split torsion theory. Then SX ⊆ X .

Proof. Remember S ∼= Στ ∼= τΣ (Definition 0.3.11). Let u = (g, h) ∈ X andconsider the region Ru in Proposition 1.4.6. Thus Sx = S(g, h) = τΣ(g, h) =τ(h− 1, g + n + 2) = (h− 2, g + n + 1) is in Ru = (x, y)|g ≤ x, h ≤ y and2 ≤ y − x ≤ n+ 1 ⊆ X .

An alternative proof of Lemma 1.4.9 is then given.

Proof. Since (X ,Y) is a split torsion theory, S−1ΣX ⊆ X by Theorem 1.3.5and SX ⊆ X by Lemma 1.4.10. Finally, by Lemma 1.3.1(i), S−1ΣX ⊆ Xand SX ⊆ X give ΣX ⊆ X .

The following lemma is true not only for the subcategory X of the finitederived category Db(mod kAn), but simply of any k-linear Hom finite andKrull-Schmidt triangulated category. It is suggestive of the situation whenthe notions of split torsion theories and t-structures coincide.

Lemma 1.4.11. Suppose SX ∼= X . Then the following are equivalent.

(i) S−1ΣX ⊆ X ,

(ii) ΣX ⊆ X .

55

Proof. This follows by Lemma 1.3.1(i) and Lemma 1.3.1(iii).

Remark 1.4.12. Suppose a torsion theory satisfies the condition in Lemma 1.4.11.Then the one characterized by (i) corresponds to a split torsion theory byTheorem 1.3.5, and the one characterized by (ii) corresponds to a t-structureby definition.

The following is an illustration of Lemma 1.4.11.

Example 1.4.13. Consider the finite derived category Db(k) = Db(mod kAn)where n = 1. By [15, Section 5.2], an indecomposable X in Db(k) is of theform

· · · // 0 // 0 // k // 0 // 0 // · · · ,

where k is at any i-th position.

By Lemma 0.3.6 and Lemma 0.1.9, the end terms of an Auslander-Reiten tri-angle are indecomposable. Therefore an Auslander-Reiten triangle in Db(k)is of the form

Xα // Y

β // Zγ 6=0 // ΣX,

where X and Z are indecomposable, and Σ−1Z ∼= X since (Z,ΣX) is one-dimensional and the morphism γ : Z → ΣX is non-zero. Therefore Y ∼= 0 byLemma 0.2.2(iii) and an Auslander-Reiten triangle in Db(k) is of the form

X // 0 // Zγ 6=0 // ΣX.

Together with τZ ∼= SΣ−1Z ∼= X this gives SX ∼= X.

The Auslander-Reiten quiver of Db(k) is as follows.

· · · • • • • • · · ·

Remark 1.4.14. Let (X1,Y1) be a torsion theory in Db(k) (Example 1.4.13).Then by Lemma 1.4.11, the torsion theory is a split torsion theory if andonly if it is a t-structure (c.f. Remark 1.4.12).

1.5 Anecdote

The chapter concludes with the following informal recapitulation. As usual,let k be a field. Let T be a k-linear Hom finite and Krull-Schmidt triangu-

56

lated category with translation functor Σ, Serre functor S and Auslander-Reiten translation τ ∼= SΣ−1. Let X , Y be full additive subcategories of T .They do not need to be triangulated, but are assumed to be closed underisomorphisms.

Three different types of distinguished triangles have been hitherto described.

(i) split distinguished triangles,

(ii) torsion theory triangles,

(iii) Auslander-Reiten triangles.

Torsion theory triangles can be split. On the other hand, Auslander-Reitentriangles are not split.

Lemma 1.5.1. Let (X , Y) be a torsion theory. If τ−1X ⊆ X , then there areno distinguished triangles (other than the one with all terms zero) ε where εis both a torsion theory triangle and an Auslander-Reiten triangle.

Proof. Suppose ε : x → t → y → Σx is such a triangle. Then by Theo-rem 1.3.5, ε is split, but by definition Auslander-Reiten triangles are notsplit. Alternatively, τ−1X ⊆ X implies that y is in X , but then y wouldhave to be in X ∩ Y = 0 (Remark 1.2.6(i)). Similarly, x would also have tobe 0 since τY ⊆ Y by Theorem 1.3.5.

Below are a few comparisons between torsion theory triangles and Auslander-Reiten triangles.

Comparison 1.5.2. (nature of restrictions) Let (X , Y) be a torsion theory,

and ε : uµ→ v

ν→ wξ→ Σu be a distinguished triangle in T . If ε is a torsion

theory triangle ε1, then it is defined in terms of the membership of u and win the given subcategories X and Y respectively, and the morphisms µ andν can be intuitively perceived as approximations of the canonical inclusionand the canonical projection respectively (Remark 1.2.30). On the otherhand, if ε is an Auslander-Reiten triangle ε2, then it is defined in terms ofthe morphisms µ, ν and ξ, and in the example given in Proposition 1.4.6,the (first three) objects u, v1 ⊕ v2 and w in the Auslander-Reiten triangleu→ v1 ⊕ v2 → w → are all in X .

Comparison 1.5.3. (construction) A torsion theory triangle is of the formRt → t → Lt → ΣRt when the right adjoint R and the left adjoint L exist(Property 1.2.28). An Auslander-Reiten triangle is of the form SΣ−1z →y → z → Sz when the Serre functor S exists (Lemma 0.3.10).

57

Comparison 1.5.4. (right minimal morphisms) Let (X , Y) be a torsion the-

ory, and ε : uµ→ v

ν→ wξ→ Σu be a distinguished triangle in T . If ε is an

Auslander-Reiten triangle, then ν is right minimal by Lemma 0.3.6. On theother hand, any given right minimal morphism µ : u → v in T , with µ anX -precover, can be extended to a torsion theory triangle u

µ→ v → w → Σuwith w in Y by Proposition 1.2.23. Accordingly, the occurrence of a rightminimal morphism in a distinguished triangle affects whether the distin-guished triangle is an Auslander-Reiten triangle or a torsion theory triangle.

Remark 1.5.5. (i) In Comparison 1.5.2, the example from Proposition 1.4.6is given, where the (first three) objects u, v1 ⊕ v2 and w in the(Auslander-Reiten) triangle u

α→ v1 ⊕ v2 → w → are all in X . Thiscan also be true in some distinguished triangles other than Auslander-

Reiten triangles. For example, consider the morphism α =

(α1

α2

)in

the above Auslander-Reiten triangle. The objects u and v1 are in X ,and in Section 3.2.2, the mapping cone y of the morphism α1 : u→ v1

is shown to be in X as well. The distinguished triangle u→ v1 → y →is not however an Auslander-Reiten triangle.

(ii) Let (X , Y) be a torsion theory. Suppose for each t in T , there is atorsion theory triangle x→ t→ y → Σx, where x ∈ X , y ∈ Y and themorphism h: Σ−1y → x is an X -left phantom. Then by Lemma 1.2.23,the right adjoint R : T → X of the inclusion ı : X → T is R(t) = x,i.e. x = Σ−1u where u is the mapping cone of the Y-preenvelope t→ y.The action of the right adjoint R can be seen on the Auslander-Reitenquiver, if it is possible to recognize the Y-preenvelope, the mappingcone construction as well as the action of Σ on the quiver. There ismore description about mapping cone constructions in some selectedsituations in Section 3.2.2, Section 3.3.2 and Section 3.4.3, and anillustration is given in Example 4.5.6.

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Chapter 2

Existence ofAuslander-Reiten sequencesin subcategories

This chapter is also written in the form of a paper which is accepted forpublication in the Journal of Pure and Applied Algebra ([36]).

2.1 Introduction

The reader is reminded of the notion of Auslander-Reiten triangles in sub-categories in Definition 0.3.5. The octahedral axiom of a triangulated cat-egory described in Section 0.2 is not necessarily true when it is confined toAuslander-Reiten triangles only. There will be more examples of applica-tions of the octahedral axiom in Section 3.4.3, where not all the distinguishedtriangles induced by the octahedral axiom are Auslander-Reiten triangles.This leads us to other conditions of the existence of Auslander-Reiten tri-angles, which is very much related to the existence of Auslander-Reitensequences described in this chapter.

Let X ,Y be full subcategories of a triangulated category T . The notions ofa torsion theory and torsion theory triangles are given in Definition 1.2.5.Also the notions of an X -(pre)cover and of X as (pre)covering for T in Defi-nition 1.2.8 and in Definition 1.2.9. Suppose (X ,Y) is a torsion theory. Theconnection between torsion theory triangles and Auslander-Reiten trianglesis very subtle, and it has been described in Comparison 1.5.4.

The following offers another facet of the interrelationship. Since (X ,Y) is atorsion theory, X is a precovering for T . If X is also a covering for T , then

59

given t in T , there is an X -cover of the form xα→ t which can be extended

to a torsion theory triangle xα→ t→ y → with y in Y by Proposition 1.2.23.

This is very similar to the existence of Auslander-Reiten triangles in sub-categories described in [22, Theorem 3.1]. Suppose T is a skeletally smallk-linear Hom finite triangulated category with split idempotents, and X isa full subcategory of T closed under extensions and direct summands. Thetheorem is restated as follows.

Theorem. ([22, Theorem 3.1]) Let x be in X and let u→ v → x→ be anAuslander-Reiten triangle in T . Then the following are equivalent.

(i) u has an X -cover of the form a→ u,

(ii) There is an Auslander-Reiten triangle a→ b→ x→ in X .

Therefore there is a very close and subtle relationship between the existenceof torsion theory triangles and the existence of Auslander-Reiten triangles.In this chapter, we study the existence of Auslander-Reiten sequences insubcategories of mod(Λ), where Λ is a finite-dimensional k-algebra over thefield k, based on the theory of the existence of Auslander-Reiten trianglesin subcategories developed in [22].

In the above theorem, when u is in X , (i) and (ii) are true trivially andindependently. Therefore intuitively, (i) and (ii) are two equivalent forms indisguise for measuring how far for each u /∈ X it is from X . Accordingly,X -covers and Auslander-Reiten triangles in X are different expressions ofthe subcategory X approximating T .

2.2 Notations

Let Λ be a finite-dimensional k-algebra over the field k, and Λop be theopposite algebra. The elements of Λ and Λop are the same, and given multi-plication · and elements λ1, λ2 in Λop, let λ1·λ2 = λ2λ1, where λ2λ1 is consid-ered in Λ. As usual, let mod(Λ) (resp. mod(Λop)) be the category of finitelygenerated Λ-left-modules (resp. Λop-left-modules or Λ-right-modules).

A finite-dimensional k-algebra Λ is said to be representation-finite if thenumber of isomorphism classes of indecomposable finite-dimensional Λ-left-modules is finite. Otherwise, it is said to be representation-infinite.

Let D be the usual duality functor D(−) = Homk(−, k) : mod(Λ) →mod(Λop). Given a Λ-left-module M in mod(Λ), the set Homk(M,k) is ak-vector space, and is also a Λ-right-module. Indeed given µ in Homk(M,k)

60

and λ in Λ, define (µ · λ)(m) = µ(λ ·m) for m in M . In particular, DΛ =Homk(Λ, k) is the k-linear dual of Λ which is a Λ-bi-module.

The duality functor D(−) is an exact, contravariant functor and it induces acontravariant equivalence between the categories mod(Λ) and mod(Λop), sothat the category mod(Λop) is equivalent to the opposite category of mod(Λ).

Lemma 2.2.1. Let M be in mod(Λ). Then

(i) M is indecomposable if and only if DM is indecomposable,

(ii) If M is projective (resp. injective), then DM is injective (resp. pro-jective).

Subsequently, the duality D(−) : mod(Λ) → mod(Λop) induces a dualityD(−) : mod(Λ)→ mod(Λop).

LetM be a full subcategory of mod(Λ), and remember the notions of a rightalmost split morphism and of a left almost split morphism in Definition 0.3.1.

Definition 2.2.2. (c.f. Definition 0.3.5) An exact sequence 0→ Ag→ B

f→C → 0 with A,B,C in the subcategory M is an Auslander-Reiten sequencein M if g is left almost split and f is right almost split in M.

Following the notations of [22], let K(InjΛ) be the homotopy category ofcomplexes of injective Λ-left-modules. Let T be the full subcategory ofK(InjΛ) consisting of complexes X for which each Xi is finitely generated,and where Hi(X) = 0 for i 0 and HiHomΛ(DΛ, X) = 0 for i 0.

The category T is a skeletally small k-linear Hom finite triangulated categorywith split idempotents. It is Krull-Schmidt as well. Let the suspensionfunctor Σ be the translation functor of the triangulated category T . LetC be the full subcategory of T which consists of injective resolutions ofmodules in mod(Λ). The category C is closed under extensions and directsummands. The reader can refer to [22, Remark 1.2] and [22, Lemma 4.3]for more details.

Let M be a full subcategory of mod(Λ) closed under extensions and directsummands, and let C′ consist of the injective resolutions of the M in M.The subcategories C′ and M need not be abelian.

Remark 2.2.3. (i) C is equivalent to mod(Λ), by the functor F : mod(Λ)→C which sends X in mod(Λ) to its injective resolution C in C. Similarly,C′ and M are equivalent.

(ii) By [22, Lemma 4.5], C is covering in T . If C′ is precovering in C, thenC′ is precovering in T .

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By Remark 2.2.3(i), properties in the category mod(Λ) can be passed onto the category C, and vice versa. For example, mod(Λ) is closed underextensions and direct summands if and only if C is closed under extensionsand direct summands. Also C′ is (pre)covering in C if and only if M is(pre)covering in mod(Λ).

The setup described above could be summarized in the following diagram,

M //

'

mod(Λ)

'

C′ // C

// T // K(InjΛ).

Remark 2.2.4. Intuitively, C stays away from T far enough to be abelian(mod(Λ) is abelian). On the other hand, it needs to stay close enough sothat there are C-covers of objects in T .

Finally, the bridge between Auslander-Reiten sequences and Auslander-Reiten triangles in subcategories is shown in the following lemma.

Lemma 2.2.5. 0 → X → Y → Z → 0 is an Auslander-Reiten sequencein M if and only if A → B → C → is an Auslander-Reiten triangle in C′,where A,B,C are injective resolutions of X,Y, Z respectively.

Proof. (if ) This is Lemma 0.2.9. (only if ) This is the dual of Lemma 0.2.22.Essentially, M and C′ are equivalent so that short exact sequences in Mcorrespond to distinguished triangles in C′.

2.3 Weakened notions of precovers (preenvelopes)

Let us begin this section with some notations and definitions.

Let (−)∗ be the functor HomΛ(−,Λ) : mod(Λ) → mod(Λop). Given a Λ-left-module M in mod(Λ), the set HomΛ(M,Λ) is a k-vector space, and isalso a Λ-right-module. Indeed given µ in HomΛ(M,Λ) and λ in Λ, define(µ · λ)(m) = µ(m) · λ for m in M . In particular, HomΛ(Λ,Λ) = Λ.

The functor (−)∗ is a left exact contravariant functor. However, unlikethe duality functor D(−), the functor (−)∗ does not induce an equivalencebetween the categories mod(Λ) and mod(Λop). Therefore, the functor (−)∗

is not a duality and given N in mod(Λop), there is not necessarily an M inmod(Λ) such that M∗ = N .

Lemma 2.3.1. (c.f. Lemma 2.2.1) Let M be in mod(Λ).

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(i) If M is free, then HomΛ(M,Λ) is also free,

(ii) If M is projective, then HomΛ(M,Λ) is also projective,

(iii) If M is projective, then M∗∗ ∼= M .

Let M be in M. Let

P = · · · // P2// P1

// P0// 0 // · · ·

be a projective resolution of M . The Pi do not need to be in M.

The sequenceP1

// P0//M // 0

is right exact, and the functor (−)∗ induces the following left exact sequence,

0 //M∗ // P ∗0// P ∗1 .

Since the functor (−)∗ is not an exact functor, this leads us to the followingdefinition.

Definition 2.3.2. The transpose TrM of M is defined to be the cokernelof the map P ∗0 → P ∗1 .

The transpose TrM is a k-vector space and is also a Λ-right-module inmod(Λop). The transpose does not induce a duality Tr(−) : mod(Λ) →mod(Λop), and in general not even a functor from mod(Λ) to mod(Λop)since Tr(−) : mod(Λ)→ mod(Λop) is well-defined on objects but in generalnot well-defined on morphisms.

Lemma 2.3.3. ([2, Proposition IV.1.7]) Let M be in mod(Λ). Then

(i) TrM = 0 if and only if M is projective,

(ii) If M is not projective, then M is indecomposable if and only if TrMis indecomposable.

Subsequently, the transpose Tr(−) induces a well-defined contravariant func-tor and along with it a duality Tr(−) : mod(Λ)→ mod(Λop).

Now let Pi be a projective module in mod(Λ). The tensor product DΛ⊗ΛPiis naturally equipped with the structure of a Λ-left-module. Given d in DΛ,p in Pi and λ in Λ, define λ · (d ⊗ p) = (λ · d) ⊗ p. By Lemma 0.1.4(ii),DΛ⊗Λ Pi is injective.

Since DΛ⊗Λ Pi ∼= Homk(HomΛ(Pi,Λ), k) = D(P ∗i ), we write

DΛ⊗Λ P = D(P ∗) = · · · //D(P ∗2 ) //D(P ∗1 )d1 //D(P ∗0 ) //0 // · · · .

63

Remark 2.3.4. Even though the DΛ⊗ΛPi are injective, the complex DΛ⊗ΛPis far from being injective, let alone an injective resolution.

From the exact sequence

0 //M∗ // P ∗0// P ∗1

// TrM // 0,

and the consideration that the functor D(−) preserves exactness, there isthe following exact sequence,

0 // DTrM // D(P ∗1 ) // D(P ∗0 ) // D(M∗) // 0.

Accordingly, DTrM and D(M∗) are the kernel and cokernel of the map d1.

Remark 2.3.5. (i) Even though M is inM, the module DTrM might notbe in M,

(ii) The composition DTr(−) : mod(Λ)→ mod(Λ) induces an equivalenceDTr(−) : mod(Λ) → mod(Λ) of categories with inverse equivalenceTrD(−) : mod(Λ)→ mod(Λ),

(iii) The right exact sequence P ∗0 → P ∗1 → TrM → 0 induces a left exactsequence 0 → (TrM)∗ → P ∗∗1 → P ∗∗0 . Since P ∗∗i

∼= Pi for i = 0, 1 byLemma 2.3.1(iii), there is the exact sequence 0 → (TrM)∗ → P1 →P0 → M → 0. The module (TrM)∗, which is the kernel of the mor-phism P1 → P0, is known as the second syzygy of M , denoted byΩ2M .

Now we are ready to introduce a weakened notion of anM-precover, i.e. anM-precover with error term, and discover its relationship with a C′-precover.

Definition 2.3.6. Let M and N be in M. Then ν : N → DTrM is saidto be an M-precover with error term if for all L in M, each morphisms′ : L → DTrM factors through ν up to an error term, i.e. there is amorphism ν ′ : L→ N such that νν ′ − s′ factors through f2 in the following

way: L→ D(P ∗2 )f2→ DTrM , as indicated in the following diagram.

L

s′

ν′

Nν

Σ−1D(P ∗) = //D(P ∗2 )

f2''

d2 //D(P ∗1 )d1 //D(P ∗0 ) //0 //

DTrM* j1

77

64

Remark 2.3.7. In Definition 2.3.6, there is a unique morphism f2 : D(P ∗2 )→DTrM since DTrM is the kernel of the map d1.

Similarly, consider the subcategory DM of mod(Λop). The subcategoryDM is equivalent to the opposite category of M. Let M be in M. Let

Q = · · · // Q2// Q1

// Q0// 0 // · · ·

be a projective resolution of DM .

Definition 2.3.8. Let M and N be inM. Then Df : DN → DTr(DM) issaid to be a DM-precover with error term if for all L inM, each morphismDs′ : DL → DTr(DM) factors through Df up to an error term, i.e. thereis a morphism Df ′ : DL → DN such that DfDf ′ − Ds′ factors through

D(Q∗2) in the following way: DL→ D(Q∗2)D(g2)→ DTr(DM), as indicated in

the following diagram.

DL

Ds′

Df ′

DNDf

Σ−1D(Q∗) = //D(Q∗2)

D(g2) ((

//D(Q∗1) //D(Q∗0) //0 //

DTr(DM))

66

There is also the dual version.

Definition 2.3.9. Let M and N be in M. Then f : Tr(DM)→ N is saidto be an M-preenvelope with error term if for all L in M, each morphisms′ : Tr(DM) → L factors through f up to an error term, i.e. there is amorphism f ′ : N → L such that f ′f − s′ factors through Q∗2 in the followingway: Tr(DM)→ Q∗2 → L.

Lemma 2.3.10. Let M , N be in M and ν : N → DTrM be given. Let Jin C′ be the injective resolution of N . Then there is a chain map λ : J →Σ−1D(P ∗) induced by ν as indicated in the following diagram.

J =

λ

//0

//J0

λ0

a0//J1

λ1

//J2 //

Nν

* i

77

Σ−1D(P ∗) = //D(P ∗2 )

f2''

d2 //D(P ∗1 )d1

//D(P ∗0 ) //0 //

DTrM* j1

77

65

Proof. This is Lemma 0.1.27(ii), but we shall show it explicitly here. TheJ i here do not need to be in M.

J =

λ

//0

a−1//J0

a0//

h1&& &&

J1 //

J2 //

Nν

) i

66

σN+

i1

88

Σ−1D(P ∗) = //D(P ∗2 )

f2''

d2 //D(P ∗1 )d1

//

f1%% %%

D(P ∗0 ) //0 //

DTrM*

j1

77

C j2

99

In the above diagram, let σN and C be the cokernels of i and of j1 respec-tively. Since a0a−1 = 0 and σN is the cokernel, there is a unique i1 : σN →J1 such that a0 = i1h1. Similarly, there is a unique j2 : C → D(P ∗0 ) suchthat d1 = j2f1.

Since Z0(J) = N and D(P ∗1 ) is injective, there is a morphism λ0 : J0 →D(P ∗1 ) such that λ0i = j1ν. Since f1λ0i = f1j1ν = 0, there is a uniquemorphism µ : σN → C such that µh1 = f1λ0, since C is the cokernel.Finally, there is a morphism λ1 : J1 → D(P ∗0 ) such that λ1i1 = j2µ sinceD(P ∗0 ) is injective. Hence λ1a

0 = d1λ0 and there is a chain map λ : J →Σ−1D(P ∗), which is only dependent on the morphism ν : N → DTrM .

Remark 2.3.11. Conversely, given a chain map λ : J → Σ−1D(P ∗), sinceZ0(J) = N , therefore d1λ0i = λ1a

0i = 0, and so there is a unique morphismν : N → DTrM such that j1ν = λ0i, since DTrM is the kernel of the mapd1.

Lemma 2.3.12. Let M , N be in M and ν : N → DTrM be given. Let Jin C′ be the injective resolution of N and let λ : J → Σ−1D(P ∗) be a chainmap induced by ν as indicated in the following diagram.

J =

λ

//0

//J0

λ0

ϕ0

a0//J1

λ1

ϕ1

//J2 //

Nν

* i

77

Σ−1D(P ∗) = //D(P ∗2 )

f2''

d2 //D(P ∗1 )d1

//D(P ∗0 ) //0 //

DTrM* j1

77

If λ is null homotopic, then ν factorizes as f2ϕ0i for some ϕ0.

Proof. Since λ is null homotopic, λ0 = d2ϕ0 + ϕ1a0 = j1f2ϕ

0 + ϕ1a0 forsome ϕ0, ϕ1. Hence λ0i = j1f2ϕ

0i + ϕ1a0i = j1f2ϕ0i since a0i = 0. Since

66

λ0i = j1ν (by construction), we have j1ν = j1f2ϕ0i. Since j1 is injective,

ν = f2ϕ0i.

Lemma 2.3.13. Let M , N be in M and ν : N → DTrM be given. Let Jin C′ be the injective resolution of N and let λ : J → Σ−1D(P ∗) be a chainmap induced by ν as indicated in the following diagram.

J =

λ

//0

//J0

λ0

ϕ0

a0//

h1 %% %%

J1

λ1

ϕ1

a1//J2 //

ϕ2

Nν

* i

77

σN+ i1

99

gvv

Σ−1D(P ∗) = //D(P ∗2 )

f2''

d2 //D(P ∗1 )d1

//D(P ∗0 ) //0 //

DTrM* j1

77

If ν factorizes as f2ϕ0i for some ϕ0, then the chain map λ is null homotopic.

Proof. Since λ0i = j1ν, we have λ0i = j1f2ϕ0i = d2ϕ

0i. Hence (λ0 −d2ϕ

0)i = 0. Let σN be the cokernel of i, then there is a unique g: σN →D(P ∗1 ) such that λ0 − d2ϕ

0 = gh1. Since D(P ∗1 ) is injective, there is ϕ1 :J1 → D(P ∗1 ) such that ϕ1i1 = g. Hence λ0 − d2ϕ

0 = ϕ1i1h1 so λ0 =d2ϕ

0 + ϕ1i1h1 = d2ϕ0 + ϕ1a0. Similarly, we obtain a map ϕ2 : J2 → D(P ∗0 )

such that λ1 = d1ϕ1 + ϕ2a1.

The following proposition is a slight variation of Lemma 0.1.27(ii).

Proposition 2.3.14. Let M be in M. Then DTrM has an M-precoverwith error term if and only if Σ−1D(P ∗) has a C′-precover in T .

Proof. Since Σ−1D(P ∗) need not be in C, we cannot appeal to Remark 2.2.3.We begin by giving a useful diagram.

K =

s

r

//0 //

K0

r0

//

ϕ

K1

r1

//K2 //

L

s′

ν′

* il

77

J =

λ

//0 //

J0

λ0

a0//J1

λ1

a1//J2 //

Nν

* i

77

Σ−1D(P ∗) = //D(P ∗2 )

f2''

//D(P ∗1 )d1 //D(P ∗0 ) //0 //

DTrM* j1

77

The following discussion is with reference to the diagram.

67

(only if ) Let ν : N → DTrM be an M-precover with error term. Let J bean injective resolution of N and extend ν to a chain map λ : J → Σ−1D(P ∗)by Lemma 2.3.10. Then λ is a C′-precover. First of all, J is in C′ since Nis in M. Suppose K is in C′ with a chain map s : K → Σ−1D(P ∗). Thenthere is the induced map s′ : Z0(K) = L→DTrM . Since L is inM and ν isan M-precover with error term, there is a morphism ν ′ : L → N such thatνν ′ − s′ = f2ϕil for some ϕ : K0 → D(P ∗2 ). By Lemma 2.3.10, extend ν ′ toa chain map r : K → J . By Lemma 2.3.13, λr − s is null homotopic, thatis, λr = s in T and λ is a C′-precover. (if ) Suppose λ : J → Σ−1D(P ∗) is aC′-precover. Then there is a morphism ν : Z0(J) = N → DTrM , and ν is anM-precover with error term. Suppose we are given s′ : L → DTrM whereL is in M. By Lemma 2.3.10, extend s′ to a chain map s : K → Σ−1D(P ∗)where K is an injective resolution of L. Since λ is a C′-precover, there isr : K → J such that λr = s. This r induces a homomorphism ν ′ : L → N .By Lemma 2.3.12, νν ′− s′ factorizes as f2ϕ

0i for some ϕ0 and ν is thereforean M-precover with error term.

Here we turn to study precovers in the stable category. First we need a littlelemma.

Lemma 2.3.15. Let U in mod(Λ) be a finitely generated injective Λ-module.Consider the complex D(P ∗) from Section 2.3. Then (U,D(P ∗2 ))→ (U,D(P ∗1 ))→(U,D(P ∗0 )) is exact.

Proof. First consider the case when U = DΛ. Then the sequence becomes(DΛ,D(P ∗2 )) → (DΛ,D(P ∗1 )) → (DΛ,D(P ∗0 )), which is isomorphic to thesequence (P ∗2 ,Λ) → (P ∗1 ,Λ) → (P ∗0 ,Λ) since the duality functor D(−) is acontravariant equivalence, which is the same as the sequence P ∗∗2 → P ∗∗1 →P ∗∗0 , which by Lemma 2.3.1(iii) is isomorphic to the exact sequence P2 →P1 → P0. Finally, any finitely generated injective is a direct summand in asum of copies of DΛ.

The following proposition shows that an M-precover with error term isequivalent to an M-precover in the stable category.

Proposition 2.3.16. Let N be in M. Then ν : N → DTrM is anM-precover with error term in mod(Λ) if and only if its class ν = ν +I(N,DTrM) is an M-precover in the stable category mod(Λ).

Proof. (only if ) Suppose ν : N → DTrM is an M-precover with errorterm. Then ν = ν + I(N,DTrM) is an M-precover in the stable category.Suppose we are given s′ : L → DTrM in the stable category with L in M,i.e. s′ : L → DTrM in mod(Λ). Since ν is an M-precover with error term,

68

there is ν ′ : L→ N such that νν ′−s′ = f2ψ for some ψ : L→ D(P ∗2 ). Hences′ = νν ′ − f2ψ so s′ = ν ν ′ - f2 ψ = ν ν ′, since f2 = 0.

(if ) Suppose ν is an M-precover in the stable category. Then ν is an M-precover with error term. Suppose we are given s′ : L → DTrM with L inM. Consider its class s′ in the stable category. Since ν is anM-precover inthe stable category, we have ν ′ : L→ N such that s′ = ν ν ′ for some ν ′, i.e.νν ′ − s′ factors through an injective U , say νν ′ − s′ = u2u1. We would like

νν ′ − s′, however, to factor through D(P ∗2 )f2→ DTrM instead.

Lνν′−s′ //

u1

##

DTrM

U

u2

99

D(P ∗2 )

f2

FF

Consider the morphism j1u2 : U →D(P ∗1 ), where j1 is as in Definition 2.3.6.

U

j1u2

g

yy

0

%%D(P ∗2 )

d2

// D(P ∗1 )d1

// D(P ∗0 )

Since d1j1u2 = 0, by Lemma 2.3.15, there is g such that d2g = j1u2, whichgives j1f2g = j1u2. Since j1 is injective, f2g = u2 and νν ′ − s = u2u1 =f2gu1.

Proposition 2.3.17. Let N be in M. Then Df : DN → DTr(DM)is a DM-precover with error term in mod(Λop) if and only if the classf = f+P(Tr(DM), N) is anM-preenvelope in the stable category mod(Λ).

Proof. Similar.

Proposition 2.3.18. Let N be in M. Then f : Tr(DM) → N is anM-preenvelope with error term in mod(Λ) if and only if the class Df =Df+I(DN,DTr(DM)) is a DM-precover in the stable category mod(Λop).

Proof. Similar.

69

2.4 Existence of Auslander-Reiten sequences in sub-categories

LetN be in mod(Λ). The functors ExtiΛ(−, N) and ExtiΛ(N,−), the right de-

rived functors of HomΛ(−, N) and HomΛ(N,−) respectively, are described in

Section 0.2.2. By Lemma 0.2.26(ii), the functors ExtiΛ(−,−) and ExtiΛ(−,−),

where i ≥ 0, are naturally equivalent.

Definition 2.4.1. ([18, III.1.], [45, 3.4]) Let A, B, X and X ′ be in mod(Λ).An extension of A by B in mod(Λ) is an exact sequence 0 → B → X →A → 0 in mod(Λ). Two extensions ε and ε′ are equivalent if there is thefollowing commutative diagram in mod(Λ).

ε : 0 // B // X //

∼=

A // 0

ε′ : 0 // B // X ′ // A // 0

An extension is split if it is equivalent to 0→ B → A⊕B → A→ 0.

Let E(A,B) be the set of equivalence classes of extensions of A by B.

The following lemmas are standard.

Lemma 2.4.2. ([45, Lemma 3.4.1]) Let A and B be in mod(Λ). If E(A,B) =0, then every extension of A by B is split.

Lemma 2.4.3. ([18, Theorem III.2.4], [18, Proposition IV.7.1], [45, Theo-rem 3.4.3]) Let A and B be in mod(Λ), and consider the extension ε : 0→B → X → A → 0 in mod(Λ). The functors ExtiΛ(A,−) induce the exact

sequence (A,X) → (A,A)∂→ Ext1(A,B) in mod(Λ). Then θ : E(A,B) →

Ext1(A,B), given by θ(ε) = ∂(idA), is a one-one correspondence, in whichthe split extension corresponds to the element 0 in Ext1(A,B).

Lemma 2.4.4. Let ε : 0 → B → X → A → 0 be an extension in mod(Λ).If either A is projective or B is injective, then the extension ε is split.

Proof. This is true by Lemma 0.2.26(i), Lemma 2.4.2 and Lemma 2.4.3.

Let us reminisce the following with regard to Auslander-Reiten sequences inmod(Λ).

Lemma 2.4.5. ([2, Theorem V.1.16]) Let 0 → Ag→ B

f→ C → 0 and

0 → A′g′→ B′

f ′→ C ′ → 0 be two Auslander-Reiten sequences in mod(Λ).Then the following are equivalent.

70

(i) A ∼= A′,

(ii) C ∼= C ′,

(iii) The sequences are isomorphic, which is to say there is a commutativediagram,

0 // Ag //

∼=

Bf //

∼=

C

∼=

// 0

0 // A′g′ // B′

f ′ // C ′ // 0,

where all the vertical morphisms are isomorphisms.

Lemma 2.4.6. ([2, Proposition V.1.14]) Let 0 → Ag→ B

f→ C → 0 be anexact sequence in mod(Λ). Then the following are equivalent.

(i) The sequence is an Auslander-Reiten sequence,

(ii) The module A is indecomposable and the morphism f is right almostsplit,

(iii) The module C is indecomposable and the morphism g is left almostsplit,

(iv) A ∼= DTrC and the morphism f is right almost split,

(v) C ∼= TrDA and the morphism g is left almost split.

By Lemma 2.4.5 and Lemma 2.4.6, the existence of Auslander-Reiten se-quences in mod(Λ) implies that given an indecomposable A in mod(Λ),there is up to isomorphism a unique indecomposable C in mod(Λ) such thatDTrC ∼= A. Similarly, given an indecomposable C in mod(Λ), there is up toisomorphism a unique indecomposable A in mod(Λ) such that TrDA ∼= C.

Originally in [3, Theorem 2.4], Auslander and Smalø developed a theoryfor the existence of Auslander-Reiten sequences in subcategories of mod(Λ).Then in [28, Corollary 2.8], Kleiner gave a new proof of their existencetheorems without the use of the theory of dualizing R-varieties. These arerephrased below.

Theorem. Let C be a precovering of mod(Λ) closed under extensions anddirect summands, and let C be an indecomposable module in C such thatExt1(C, C) 6= 0 for some C in C. Then there is an Auslander-Reiten sequence0→ A→ B → C → 0 in C.

71

Theorem. Let A be a preenveloping of mod(Λ) closed under extensionsand direct summands, and let A be an indecomposable module in A suchthat Ext1(A, A) 6= 0 for some A in A. Then there is an Auslander-Reitensequence 0→ A→ B → C → 0 in A.

In this section, we strengthen Auslander-Smalø’s and Kleiner’s results byproviding necessary and sufficient conditions, based on the theory of Auslander-Reiten triangles developed in [22]. The proof here is more subtle than theproof of [22, Theorem 3.1], since we cannot replace C by C′ and T by C,because T is triangulated and C is not triangulated.

It is conceivable that the necessary condition of the existence of Auslander-Reiten sequences in subcategories provided here is weaker, because in [22,Theorem 3.1], for the subcategory C ⊆ T to have Auslander-Reiten triangles,only certain objects X in T are required to have C-covers. This naturallyleads us to give a characterization of the existence of Auslander-Reiten se-quences in subcategories not in terms of a global condition but in terms ofa condition which describes locally at the level of objects.

Now we state the existence theorem for (right) Auslander-Reiten sequencesin subcategories.

Theorem 2.4.7. LetM be a subcategory of mod(Λ) closed under extensionsand direct summands, and let M be an indecomposable module in M suchthat Ext1(M, M) 6= 0 for some M in M. Then the following are equivalent.

(i) DTrM has an M-precover in the injective stable category mod(Λ),

(ii) There is an Auslander-Reiten sequence 0→ X → Y →M → 0 in M.

Proof. (i)⇒ (ii): Let P and C be a projective and an injective resolution ofM . By Proposition 2.3.16, DTrM has an M-precover with error term, andthen by Proposition 2.3.14, Σ−1D(P ∗) has a C′-precover. By [22, Theorem4.6], there is an Auslander-Reiten triangle A→ B → C → in C′, and finallyan Auslander-Reiten sequence 0 → H0(A) → H0(B) → H0(C) → 0 in M,where M is retrieved through the isomorphism H0(C) ∼= M (Lemma 2.2.5).

(ii) ⇒ (i): Let P and C be a projective and an injective resolution of M .Following the argument in Theorem 4.6 in [22], there is an Auslander-Reitentriangle Σ−1D(P ∗) → Y ′ → C → in T . Since there is an Auslander-Reitensequence 0 → X → Y → M → 0 in M, by Remark 2.2.5, there is anAuslander-Reiten triangle A → B → C → in C′. By Theorem 3.1 in [22],Σ−1D(P ∗) has a C′-precover. By Proposition 2.3.14, DTrM has an M-precover with error term. Finally by Proposition 2.3.16, DTrM has anM-precover in the stable category mod(Λ).

72

Remark 2.4.8. (i) In the proof of Theorem 2.4.7, (i)⇒ (ii), the Auslander-Reiten sequence not only exists, but also takes the form 0→ H0(A)→H0(B)→ H0(C)→ 0 in M.

(ii) In the proof of Theorem 2.4.7, (ii)⇒ (i), the Auslander-Reiten triangleΣ−1D(P ∗) → Y ′ → C → in T induces the exact sequence · · · → 0 →H0(Σ−1D(P ∗)) → H0(Y ′) → M → H1(Σ−1D(P ∗)) → · · · in mod(Λ)by Lemma 0.2.9, but does not render any Auslander-Reiten sequencesin mod(Λ).

Remark 2.4.9. (i) The necessary condition (i) in Theorem 2.4.7 is weakerthan that of Auslander-Smalø and Kleiner in two senses. Firstly, onlyprecovers in the stable category are considered, and secondly, onlyprecovers of the DTrM in mod(Λ) with M in M are considered. AsM is precovering in mod(Λ) if and only if C′ is precovering in C,correspondingly we do not require as much as C′ to be precovering inC either.

(ii) In the proof of Theorem 2.4.7, (ii)⇒ (i), even though D(P ∗)→ ΣY ′ →ΣC → is also an Auslander-Reiten triangle in T , the object ΣC is notnecessarily in C since C is not triangulated.

Before we give the existence theorem for (left) Auslander-Reiten sequencesin subcategories, we need the following.

Lemma 2.4.10. Let L be a subcategory of mod(Λ). Then the following areequivalent.

(i) 0→ X → Y → Z → 0 is an Auslander-Reiten sequence in L,

(ii) 0→ DZ → DY → DX → 0 is an Auslander-Reiten sequence in DL.

Proof. Please refer to the remark after [2, Proposition V.1.13].

Proposition 2.4.11. Let L be a subcategory of mod(Λ). Denote DL byM.Let L be in L. Then the following are equivalent.

(i) DL has an M-precover in the injective stable category mod(Λop),

(ii) L has an L-preenvelope in the projective stable category mod(Λ).

Proof. This can be shown by standard arguments. However, we shall onlyconsider the special case L = TrDL′ for some L′ in L, since this is whatis needed in Theorem 2.4.13. Let N be in L. Then by Proposition 2.3.17,f = f + P(TrDL′, N) is an L-preenvelope in the projective stable categorymod(Λ) if and only if Df : DN → DTrDL′ is a DM-precover with error

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term in mod(Λop). This is equivalent to saying f : TrDL′ → N is anL-preenvelope with error term in mod(Λ), which by Proposition 2.3.18 isequivalent to saying Df = Df + I(DN,DTrDL′) is an M-precover in theinjective stable category mod(Λop).

Proposition 2.4.12. Let L be a subcategory of mod(Λ) closed under exten-sions and direct summands. Denote DL by M. Let L be an indecomposablemodule in L such that Ext1(DL,DL) 6= 0 for some L in L. Then the fol-lowing are equivalent.

(i) DTrDL has an M-precover in the injective stable category mod(Λop),

(ii) There is an Auslander-Reiten sequence 0 → DA → DB → DL → 0in M, where A and B are in L.

Proof. Since M = DL, and L is closed under extensions and direct sum-mands,M is a subcategory of mod(Λop) closed under extensions and directsummands. Since L is an indecomposable module in L, DL is an indecom-posable module in M. The rest follows from the right module version ofTheorem 2.4.7.

Finally, we give the dual of Theorem 2.4.7.

Theorem 2.4.13. Let L be a subcategory of mod(Λ) closed under extensionsand direct summands, and let L be an indecomposable module in L such thatExt1(L, L) 6= 0 for some L in L. Then the following are equivalent.

(i) TrDL has an L-preenvelope in the projective stable category mod(Λ),

(ii) There is an Auslander-Reiten sequence 0→ L→ B → A→ 0 in L.

Proof. Let M = DL. By Proposition 2.4.11, TrDL has an L-preenvelopein mod(Λ) if and only if DTrDL has an M-precover in mod(Λop). AlsoExt1(L, L) 6= 0 if and only if Ext1(DL,DL) 6= 0. Hence the result followsfrom Proposition 2.4.12, with the help of Lemma 2.4.10.

Remark 2.4.14. Theorem 2.4.13 can also be shown directly from [22, Theo-rem 3.2].

Remark 2.4.15. Theorem 2.4.7 and Theorem 2.4.13 together give back theresults by Auslander-Smalø and by Kleiner. For example, suppose M isprecovering in mod(Λ). Then given M inM, the module DTrM has anM-precover in mod(Λ) and then anM-precover in the injective stable categorymod(Λ). Finally, by Theorem 2.4.7, there is an Auslander-Reiten sequence0→ X → Y →M → 0 in M.

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2.5 An example

In this section, examples of Theorem 2.4.7 and Theorem 2.4.13 are given.First, we need a little lemma.

Lemma 2.5.1. Let M be a full subcategory of mod(Λ) closed under directsums and direct summands. Let M be in mod(Λ). Suppose there are onlyfinitely many indecomposable modules inM with non-zero maps to M . ThenM has an M-precover in mod(Λ).

Proof. Since there are only finitely many indecomposable modules in Mwith non-zero maps to M , let them be denoted by K1,K2, . . . ,Kn.

(i) Suppose n = 1.

If (K1,M) is one-dimensional with basis κ, then K1κ→M is anM-

precover. Let a morphism m : K ′ → M , with K ′ an indecomposablein M, be given.

K ′

m

K1κ //M

If K ′ 6= K1, then m is zero. If K ′ = K1, then m factors through κ bya scalar multiple of κ. On the other hand, if (K1,M) is q-dimensionalwith basis κ1, κ2, . . . , κq, then K1 ⊕K1 ⊕ . . .⊕K1︸︷︷︸

q

κ→ M is an M-

precover, where κ =(κ1 κ2 . . . κq

).

(ii) Suppose n = 2.

If (K1,M) has basis κ11, κ

12, . . . , κ

1s and (K2,M) has basis κ2

1, κ22, . . . , κ

2t ,

thenK1 ⊕K1 ⊕ . . .⊕K1︸︷︷︸s

⊕K2 ⊕K2 ⊕ . . .⊕K2︸︷︷︸t

κ→M is anM-precover,

where κ =(κ1

1 κ12 . . . κ1

s κ21 κ2

2 . . . κ2t

). Let a morphism m :

K ′ → M , with K ′ an indecomposable in M, be given. Without lossof generality, assume K ′ = K1.

K1

m

g

ttK1 ⊕K1 ⊕ . . .⊕K1 ⊕K2 ⊕K2 ⊕ . . .⊕K2

κ //M

75

If m = α1κ11 +α2κ

12 + . . .+αsκ

1s, then m can be factorized as m = κg,

where g =

α1

α2...αs00...0

. Thus M has an M-precover in mod(Λ).

The case of n > 2 is similar.

In Lemma 2.5.1, an M-precover is given by construction. For example, incase (ii), K1⊕K1⊕ . . .⊕K1⊕K2⊕K2⊕ . . .⊕K2

κ→M is anM-precover.This is not the minimal construction when, for example, each κ1

i factorsthrough κ2

j , some j, in which case K2⊕K2⊕ . . .⊕K2 →M will be sufficientto be an M-precover.

Example 2.5.2. Let Λ be a representation-infinite hereditary algebra. LetM be any full subcategory of mod(Λ) which consists of postprojective mod-ules, which is closed under extensions and direct summands, see [1, Def-inition VIII.2.2]. Let M be an indecomposable module in M such thatExt1(M,M) 6= 0 for some M in M. Then there is an Auslander-Reitensequence 0→ X → Y →M → 0 in M.

Proof. By [1, Lemma VIII.2.5], there are only finitely many indecompos-able modules in M which have non-zero maps to DTrM . Therefore byLemma 2.5.1, DTrM has anM-precover in mod(Λ) and then anM-precoverin the stable category mod(Λ). The existence of the Auslander-Reiten se-quence 0→ X → Y →M → 0 in M follows from Theorem 2.4.7.

Remark 2.5.3. Dually, let L be the full subcategory of mod(Λ) consistingof preinjective modules over Λ and is closed under extensions and directsummands. Let L be an indecomposable module in L such that Ext1(L, L) 6=0 for some L in L. Then the existence of the Auslander-Reiten sequence0→ L→ Y → X → 0 in L follows from Theorem 2.4.13.

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Chapter 3

Quotients of certaintriangulated categories asderived categories

3.1 Introduction

Let us begin with the notion of a pretriangulated category. This is an ad-ditive category with an endofunctor σ and a class of diagrams of the formx→ y → z → σx known as distinguished right triangles, and another endo-functor ω and a class of diagrams of the form ωz → x → y → z known asdistinguished left triangles. The distinguished right triangles have to satisfythe axioms of a triangulated category, except that σ does not need to bean equivalence. Similarly for the distinguished left triangles. It is worth ac-knowledging that (σ, ω) is an adjoint pair of functors, which is reminiscentof the intuition that adjoints are approximations of inverse functors.

Let k be a field and T be a k-linear triangulated category with translationfunctor Σ and Serre functor S, which is Hom finite and Krull-Schmidt. Let Xbe a subcategory of T which is both precovering and preenveloping, and letTX be the quotient category defined in [23, Introduction]. By [23, Theorem1.2], the quotient category TX is pretriangulated. The quotient categoryTX is triangulated if and only if τX = X , where τ is the Auslander-Reitentranslation, see [23, Theorem 2.3]. This means the subcategory X is a unionof orbits of τ .

The definition of a Frobenius category is given in [14]. A subcategory of atriangulated category is thick if it is triangulated and closed under directsummands. A thick subcategory generated by X, where X is an object ofT , is denoted by thick(X). A triangulated category T is of algebraic origin

77

if it is equivalent, as a triangulated category, to the quotient category EP ofa k-linear Frobenius category E , where P is the class of projective objectsof E , see [23, Section 3]. By [23, Theorem 3.2], if T is of algebraic origin,then TX is also of algebraic origin.

Let Λ be a finite-dimensional k-algebra over the field k. A compact derivedcategory Dc(Λ) is the full subcategory of the derived category D(Λ) con-sisting of compact objects, i.e. those finitely built from the algebra Λ usingsuspensions, distinguished triangles, direct sums and direct summands. If Λis Noetherian and of finite global dimension, then the compact derived cat-egory Dc(Λ) is equivalent to the finite derived category Db(Λ). Let Dc(A)be the compact derived category generated by A.

In this chapter, quotients of certain triangulated categories are triangulatedand are in addition derived categories, appealing to the following theoremwhich is a slight variation of [40, Theorem 6.4] and of [26, 3.3].

Theorem 3.1.1. Let k be a field and U be a k-linear triangulated categoryof algebraic origin such that U = thick(X) for some X in U . Let Σ be thetranslation functor of the triangulated category U and let A = EndU (X).Suppose HomU (X,ΣiX) = 0 for i 6= 0. Then there is an equivalence of

triangulated categories f : U '→Dc(A) where f(X) = A.

3.2 The finite derived category Db(mod kAn)

In this section, let T be the finite derived category Db(mod kAn) describedin Section 1.4. The category T is a k-linear Hom finite triangulated category.It is Krull-Schmidt as well. The Auslander-Reiten quiver of Db(mod kAn)is reproduced here, with the coordinate system described in Section 1.4.

•

•

•tn

•

•

•

??

•

??

•

??

•

??

•

??

. . . •

??

•

??

•

??

•

??

• . . .

•

??

•

??

•

??

•

??

•

??

•

??

•

??

•

??

•

??

•

•t1

??

•

??

•

??

•

??

•

??

Given an indecomposable object (i, j) of Db(mod kAn), the action of Σ is

78

given by Σ(i, j) = (j−1, i+n+2), and the action of τ by τ(i, j) = (i−1, j−1)(Section 1.4).

The following is an example of Lemma 1.4.1 and the narration after it, andis rewritten below to remind the reader.

Lemma 3.2.1. Let x and y be indecomposable objects of Db(mod kAn).Then by [14, 4.6], any non-zero morphism f from x to y is a linear combi-nation of morphisms, written f = Σαifi, where the αi are scalars and thefi : x → y are compositions of irreducible morphisms. The finite derivedcategory Db(mod kAn) is also standard.

The Auslander-Reiten quiver of Db(mod kAn) contains the following localconfiguration throughout, which gives rise to Auslander-Reiten triangles andcommutativity relations.

Configuration 3.2.2.

bt

as??

q

d

cr

??

In Configuration 3.2.2, since a → b ⊕ c → d → is an Auslander-Reitentriangle, ts + rq = 0. This gives ts = −rq, which means the differentmorphisms from a to d are equal up to signs by Lemma 3.2.1.

Given an indecomposable object a of Db(mod kAn), the notions of L(a)and of R(a), and the sketches of the two respective regions, are describedin Section 1.4. Given indecomposable objects x and y of Db(mod kAn), if(x, y) is non-zero, then by Lemma 3.2.1 and by virtue of the little remarkafter Configuration 3.2.2, (x, y) is one-dimensional.

With the following lemma we end the section.

Lemma 3.2.3. Let a, b and c be indecomposable objects of Db(mod kAn).Let c be in R(a), and let f : a → b and g : b → c be non-zero morphisms.Then the composition gf : a→ c is non-zero.

Proof. Assume gf is zero. Let h be a non-zero morphism in (a, c). Since(a, c) is non-zero, it is one-dimensional. Thus the morphism h is a composi-tion of irreducible morphisms. Furthermore, the morphism h can be writtenh = αgf for some scalar α and some non-zero g : b → c by the commu-tativity relation described above. Since (b, c) is one-dimensional, g = βgfor some scalar β. However, this would give h = αβgf = 0, which is acontradiction.

79

3.2.1 Lemmas

Consider the indecomposable objects in a band of width n′ vertices alongthe top of the Auslander-Reiten quiver of T . Henceforth, let X be add ofthem.

Lemma 3.2.4. In the following sketch, the indecomposable objects of X liein the region bounded by and including the two upper lines. Let m be anindecomposable object in T . Then m has an X -cover g : x→ m. Similarly,m has an X -envelope.

Proof. If m is in X , then the lemma is trivial. The following sketch showsthe case where m is not in X .

b

a x

L(m)

L(x) m

Let x be an object in X which lies on the bottom line of the upper band,and is on the upper right hand boundary of the region L(m). Then thenon-zero morphism g : x → m is an X -precover of m. Indeed consider anindecomposable y in X with a non-zero morphism h : y → m. Then y is inthe region abx which is the intersection of X and L(m). Since the region abxis inside L(x), there is a non-zero morphism f : y → x. By Lemma 3.2.3,the composition gf is non-zero. Since h and gf are both in (y,m) which isone-dimensional, h = k(gf), where k is a scalar, and so h factors through g.

Furthermore, g is an X -cover. Suppose ϕ : x → x is a morphism such thatgϕ = g. Then again since (x, x) is one-dimensional, ϕ = α · idx where αis a scalar, so gϕ = g gives α · g = g but g 6= 0 so α = 1. Therefore ϕ isinvertible. The dual argument gives an X -envelope.

Remark 3.2.5. In Lemma 3.2.4, the region abx, which is the intersection ofX and L(m), has only finitely many objects. Therefore an X -precover of mcan be obtained in the same fashion as in Lemma 2.5.1. The above showshow an X -precover can be strengthened into an X -cover.

80

By Lemma 3.2.4 and [23, Theorem 1.2], the quotient category TX is pretri-angulated. Since evidently τX = X , therefore TX is in addition triangulatedby [23, Theorem 2.3]. By [23, Theorem 3.2], the Auslander-Reiten quiver ofTX is obtained by deleting the upper n′ lines of vertices and the incident ar-rows from the Auslander-Reiten quiver of T . Suppose the Auslander-Reitenquiver of TX is of width m vertices.

Definition 3.2.6. Let a be an indecomposable object in TX . Similarly toDefinition 1.4.2, let LX (a) be the set of indecomposable objects with non-zero morphisms to a in TX . Dually, let RX (a) be the set of indecomposableobjects to which there are non-zero morphisms from a in TX .

Lemma 3.2.7. Suppose the Auslander-Reiten quiver of T lies in the fol-lowing sketch in the region bounded by y − x = 2 and y − x = n + 1, andthe indecomposable objects of X form the upper band, which is the regionbounded by y − x = m+ 2 and y − x = n+ 1. The Auslander-Reiten quiverof TX lies in the lower band, which is the region bounded by y − x = 2 andy − x = m+ 1. Here the lower band is of width m vertices.

d′ dy−x=n+1

wy−x=m+2

c′ v′ v cy−x=m+1

u′ •s u

LX (a) a RX (a)

b′ by−x=2

Then LX (a) is the region ab′u′v′ and RX (a) is the region abuv.

Proof. Since R(a) is the region abcd (Sketch 1.4.3), RX (a) is at most theregion abuv, since any path from a to an object in R(a) outside abuv canbe written as a path which factors through the object w in X , by virtue ofthe little remark after Configuration 3.2.2. Now let s be an object withinthe region abuv. Since s is in R(a), there is a non-zero morphism f : a→ sin T , and the corresponding morphism f : a → s in TX is non-zero in TX .Suppose it is not. Then f is a morphism in T which factors through a (notnecessarily indecomposable) object t in X , i.e. f factorizes as f : a→ t→ sin T . By Lemma 3.2.4, a has the X -preenvelope a→ w, therefore f furtherfactorizes as f : a → w → t → s in T . However, s is not in R(w), which

81

contradicts f non-zero. This gives f non-zero in TX . Also Lemma 3.2.3 isimplicitly used throughout. The situation for LX (a) is similar.

The following lemma shows how a triangulated subcategory can be recoveredfrom a finite set of indecomposable objects by virtue of the Auslander-Reitentriangles alone.

Lemma 3.2.8. The following sketch shows some indecomposable objects in

TX . Consider t =

m⊕i=1

ti in TX . Then the thick subcategory thick(t) of TX ,

generated by t, is in fact equal to TX .

tm um

· · · · · ·

t2 u2

t1 u1

Proof. This makes sense since thick(t) is triangulated by definition and TXis triangulated by [23, Theorem 2.3]. For 1 ≤ i ≤ m, ti is in thick(t), sincethick(t) is closed under direct summands. By [23, Theorem 3.2], TX hasAuslander-Reiten triangles, and they can be read off from the Auslander-Reiten quiver. Since t1 → t2 → u1 → is an Auslander-Reiten triangle, u1

is in thick(t). Similarly, for 2 ≤ i ≤ m − 1, ti → ti+1 ⊕ ui−1 → ui →is an Auslander-Reiten triangle, and so ui is in thick(t). Finally, sincetm → um−1 → um → is an Auslander-Reiten triangle, um is in thick(t).Subsequently, all the ui immediately next to the ti on the right hand sideare in thick(t). By induction, all the indecomposable objects on the righthand side of ti are in thick(t). By a mirror argument, all the indecompos-able objects on the left hand side of ti are in thick(t). It follows that allthe indecomposable objects of TX are in thick(t), and thick(t) is equal toTX .

Lemma 3.2.9. Consider t =

m⊕i=1

ti in TX in the following sketch. Then

EndTX (t) ∼= EndT (t).

82

y

x

tm

tjti

t1

Proof. The idea is that for 1 ≤ i, j ≤ m, there are no non-zero morphismsh : ti → tj in T which factor through an object in X . Assume i ≤ jsince it is apparent that tj is not in R(ti) for j < i. Suppose there isa non-zero h : ti → tj such that h factors through an object y in X ash : ti → y → tj . In the above sketch, y is assumed to be indecomposable,though this assumption is not necessary. By Lemma 3.2.4, tj has an X -coverx→ tj , hence the morphism h further factorizes as h : ti → y → x→ tj . Butx is above and behind tj in the quiver, so that x is not in R(ti). Thereforethe only morphism from ti to tj factoring through an object in X is the zeromorphism, and the isomorphism exists.

Remark 3.2.10. The path algebra kAn is isomorphic to its opposite alge-

bra (kAn). Suppose the quiver An is x1f1→ x2 → · · · → xn−1

fn−1→ xn,then an isomorphism ϕ could be obtained by ϕ(x1) = xn, ϕ(x2) = xn−1,ϕ(f1) = fn−1, and so on. The distinguishment between kAn and (kAn) isnevertheless made explicitly, since this is not true for an arbitrary algebra.

Lemma 3.2.11. Let k be a field. Suppose that in a k-linear category S,objects p1, . . . , pm satisfy

(pi, pj) ∼=k for i ≤ j,0 otherwise.

(A)

Suppose the non-zero morphisms pi → pj and pj → pl compose to a non-zero

morphism pi → pl. Then EndS(S) ∼= (kAm) where S =m⊕i=1

pi.

Proof. Let h be an endomorphism in EndS(S),

h : p1 ⊕ p2 ⊕ · · · ⊕ pm −→ p1 ⊕ p2 ⊕ · · · ⊕ pm.

83

Then h can be viewed as a matrix,

ρ11 ρ12 · · · ρ1m

ρ21 ρ22 · · · ρ2m

. . .

ρm1 ρm2 · · · ρmm

,

where the ρij are morphisms ρij : pj → pi for 1 ≤ i, j ≤ m.

By the given condition (A) on the (pi, pj), the above matrix has the form ofa lower triangular matrix,

k11 0 0 · · · 0k21 k22 0 · · · 0k31 k32 k33 · · · 0

. . .

km1 km2 km3 · · · kmm

,

where each kji for 1 ≤ i ≤ j ≤ m takes a value in the field k. Thereforethe endomorphism algebra EndS(S) is isomorphic to the following lowertriangular m×m matrix algebra

M =

k 0 · · · 0k k · · · 0

. . .

k k · · · k

.

On the other hand, by [1, Lemma II.1.12], the path algebra kAm is alsoisomorphic to the lower triangular m×m matrix algebra M . This gives theisomorphisms EndS(S) ∼= kAm ∼= (kAm) (Remark 3.2.10).

Corollary 3.2.12. Consider the categories T , X and TX as usual, and the

object t =m⊕i=1

ti as in Lemma 3.2.8. Then EndTX (t) ∼= EndT (t) ∼= (kAm).

Proof. The first isomorphism exists by Lemma 3.2.9. The existence of thesecond isomorphism follows by identifying the ti with the pi in Lemma 3.2.11,with the help of Lemma 3.2.3.

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3.2.2 Mapping cone construction

In the following sketch, the Auslander-Reiten quiver of T lies in the regionbounded by y − x = 2 and y − x = n + 1, and the indecomposable objectsof X form the upper band, the region bounded by y − x = m + 2 andy − x = n+ 1. The Auslander-Reiten quiver of TX lies in the lower band ofwidth m vertices, the region bounded by y − x = 2 and y − x = m+ 1.

gy−x=n+1

xpy−x=m+2

ry−x=m+1c

s u

a

p b qy−x=2

Let i and j be two fixed integers. Suppose the object a has coordinates(i, j). The coordinates of some other objects are listed: p = (i, i + 2),xp = (i, i+m+2), g = (i, i+n+1), q = (i+n−1, i+n+1), s = (j−2, i+m+1).Modules are identified with their coordinates. R(a) is the region agcb, andRX (a) is the region arsb (Lemma 3.2.7).

By Lemma 3.2.4, (i, j) has the X -envelope (i, j) → (i, i + m + 2). A copyof the module category mod(kAn) can be placed inside Db(mod kAn) insuch a way that its Auslander-Reiten quiver corresponds to the region pgq.Then the modules on the line x = i may be perceived as projective modulesin mod(kAn). Perceiving the projective modules as representations of thequiver An, the projective module (i, j) becomes

0→ 0→ . . .→ 0→ k → k → . . .→ k,

where there are j − i− 1 copies of k at the end, and the projective module(i, i+m+ 2) becomes

0→ 0→ . . .→ 0→ k → k → k → . . .→ k,

where there are m+ 1 copies of k at the end.The cokernel of the embedding (i, j) → (i, i+m+ 2) is (i, i+m+ 2)/(i, j),which, when perceived as a representation, becomes

0→ . . .→ 0→ k → . . .→ k → 0→ . . .→ 0,

85

where j − i − 1 copies of k are removed from the end. Correspondingly, inthe diagram, the cokernel lies in the same descending line as (i, i+m+ 2),but it is j− i−1 steps down the line. Therefore the cokernel has coordinates(i+ (j − i− 1), i+m+ 2) = (j − 1, i+m+ 2), which is denoted by u in thediagram.

There is an alternative way of realizing the mapping cone (c.f. Section 3.3.2),and the reader will be reminded again in Remark 3.3.21. An example isdescribed in the Appendix (Lemma A.3.1 and Lemma A.3.2).

Using the same notation as in [23, Theorem 2.3], let σ be the translationfunctor of the triangulated category TX . The construction of the functor σis described in [23, Setup 1.1].

Lemma 3.2.13. For i+2 ≤ j′ ≤ i+m+1, we have σ(i, j′) = (j′−1, i+m+2).

Proof. By the above description, given the X -envelope (i, j′)→ (i, i+m+2)of (i, j′), extend it to the short exact sequence 0→ (i, j′)→ (i, i+m+ 2)→(j′−1, i+m+2)→ 0 in the module category mod(kAn). By Lemma 0.2.29,this induces a distinguished triangle in the derived category T , which isin the form of the distinguished triangle (2) in [23, Setup 1.1] when M isreplaced by (i, j′). Hence σ(i, j′) = (j′ − 1, i+m+ 2).

The special case when n = 6 and m = 4 is given in the following diagram.The action of σ is σ(ti) = si for 1 ≤ i ≤ 4.

Diagram 3.2.14.

•t6

??

??

??

•t5

??

•

??

??

??

??

??

•t4??

•s1

??

•

??

??

??

??

•t3

??

•

??

•s2

??

•

??

??

??

•t2

??

•

??

•

??

•s3

??

•

??

??

•t1??

•

??

•

??

•

??

•s4??

•

??

Corollary 3.2.15. For i+ 2 ≤ j′ ≤ i+m+ 1 and N ≥ 0 an integer,

σN (i, j′) =

(i+ N

2 (m+ 1), j′ + N2 (m+ 1)) for N even,

(j′ + N−12 (m+ 1)− 1, i+ N+1

2 (m+ 1) + 1) for N odd.

Proof. Immediate from Lemma 3.2.13 by induction.

86

For example, for i+2 ≤ j′ ≤ i+m+1, we have σ2(i, j′) = (i+m+1, j′+m+1).

Since the translation functor σ is an autoequivalence, there is also an explicitdescription of σ−1.

Lemma 3.2.16. For j −m− 1 ≤ i′ ≤ j − 2, we have σ−1(i′, j) = (j −m−2, i′ + 1).

Proof. Immediate from Lemma 3.2.13.

The special case when n = 6 and m = 4 is given in the following diagram.The action of σ−1 is σ−1(ti) = ui for 1 ≤ i ≤ 4.

Diagram 3.2.17.

•t6

??

??

??

??

??

•t5

??

•

??

u1

??

??

??

??

•t4

??

•

??

•

??

u2

??

??

??

•t3

??

•

??

•

??

??

u3

??

??

•t2

??

•

??

•

??

•

??

??

u4

??

•t1??

•

??

•

??

•

??

Corollary 3.2.18. For j −m− 1 ≤ i′ ≤ j − 2 and N ≥ 0 an integer,

σ−N (i′, j) =

(i′ − N

2 (m+ 1), j − N2 (m+ 1)) for N even,

(j − N+12 (m+ 1)− 1, i′ − N−1

2 (m+ 1) + 1) for N odd.


Lemma 3.2.19. Let N ≥ 1 be an integer. Consider t =

i+m+1⊕j′=i+2

(i, j′) in TX .

Then HomTX (t, σN t) = 0.

Proof. Suppose N = 1. The action of σ is as given in Lemma 3.2.13. Fori+2 ≤ j′ ≤ i+m+1, the set RX (i, j′) is the region bounded by (i, j′), (i, i+m + 1), (j′ − 2, i + m + 1) and (j′ − 2, j′) (Lemma 3.2.7). The idea is thatfor i + 2 ≤ j′ ≤ i + m + 1, there are no non-zero morphisms from (i, j′)to (l, i + m + 2), where i + 1 ≤ l ≤ i + m. This is apparent because fori + 1 ≤ l ≤ i + m, the object (l, i + m + 2) lies outside RX (i, j′), which isbounded by y = i+m+ 1. Suppose N > 1. Then by similar considerations,the result is immediate from Corollary 3.2.15.

87

Lemma 3.2.20. Let N ≥ 1 be an integer. Consider t =i+m+1⊕j′=i+2

(i, j′) in TX .

Then HomTX (t, σ−N t) = 0.

Proof. Similar.

3.2.3 Theorem

The following is the main theorem of the section.

Theorem 3.2.21. Consider the categories T , X and TX as usual, and the

object t =

m⊕i=1

ti as in Lemma 3.2.8. Then there is an equivalence of trian-

gulated categories f : TX'→Db(mod (kAm)) where f(t) = (kAm).

Proof. Let U and X in Theorem 3.1.1 be TX and t respectively. By Corol-lary 3.2.12, A ∼= (kAm). Let us take Σ in Theorem 3.1.1 to be the transla-tion functor σ of the triangulated category TX . Then the result follows fromLemma 3.2.8, Lemma 3.2.19 and Lemma 3.2.20.

3.3 The cluster category of Dynkin type A∞

In this section, let T be the cluster category D of Dynkin type A∞ describedin [19, Section 1]. The category D is a k-linear Hom finite triangulatedcategory. It is Krull-Schmidt as well.

The Auslander-Reiten (AR) quiver of D is ZA∞, endowed with the samecoordinate system described in [19, Remark 1.4].

...

...

...

...

· · ·

??

(−4, 1)

??

(−3, 2)

??

(−2, 3)

??

· · ·

(−4, 0)

??

(−3, 1)

??

(−2, 2)

??

(−1, 3)

??

· · ·

??

(−3, 0)

??

(−2, 1)

??

(−1, 2)

??

· · ·

(−3,−1)

??

(−2, 0)

??

(−1, 1)

??

(0, 2)

??

Since D is 2-Calabi-Yau, its Serre functor is S = Σ2 and the Auslander-Reiten translation τ is τ = SΣ−1 = Σ, where Σ is the translation functor of

88

D. In terms of coordinates, the action of Σ = τ is Σ(m,n) = (m− 1, n− 1),see [19, Remark 1.4].

The following lemma describes the morphism spaces between indecompos-able objects of D. Its significance is in giving the picture of the categoryD without a priori knowledge of how it originated and was derived. Let xbe an indecomposable object of D. The regions H−(x) and H+(x) in theAuslander-Reiten quiver of D, described in [19, Definition 2.1], are sketchedas follows.

H−(x) H+(x)

Σx x Σ−1x

We write H(x) = H−(x) ∪H+(x).

Lemma 3.3.1. ([19, Corollary 2.3]) Let x and y be indecomposable objectsof D. Then the following are equivalent.

(i) (x, y) 6= 0,

(ii) (x, y) = k,

(iii) y ∈ H(Σx),

(iv) x ∈ H(Σ−1y).

In Lemma 3.3.1, there are two different types of non-zero morphisms goingfrom x to y. Those morphisms with y in H+(x) are said to be forwardmorphisms and those morphisms with y in H−(x) are said to be backwardmorphisms.

Remark 3.3.2. (i) If f is a backward morphism, then it can always bewritten f = gf ′ where f ′ is a backward morphism and g is a forwardmorphism (Lemma 3.3.4). Therefore backward morphisms are notirreducible (can be decomposed indefinitely), and so they do not arisein the Auslander-Reiten quiver.

(ii) The analogue of Lemma 3.2.1 for D is false in view of the occurrenceof the backward morphisms.

With the following lemmas we end the section.

89

Lemma 3.3.3. Let a, b and c be indecomposable objects of D. Then

(i) b is in H+(Σa) if and only if Sa is in H−(Σb),

(ii) c is in H−(Σa) if and only if Sa is in H+(Σc).

Proof. This is [19, Lemma 2.6].

Lemma 3.3.4. Let a, b and c be indecomposable objects of D. Suppose band c are in H−(Σa) and c is in H+(Σb). Consider the non-zero morphism

f : b→ c. Then each morphism a→ c factors as a→ bf→ c.

Proof. This is [19, Lemma 2.7].

Lemma 3.3.5. (c.f. Lemma 3.2.3)

(i) Let a, b and c be indecomposable objects of D. Suppose b and c arein H+(Σa) and c is in H+(Σb). Consider the non-zero morphismsf : a→ b and g : b→ c. Then the composition gf : a→ c is non-zero.

(ii) Let a, b and c be indecomposable objects of D. Suppose b and c arein H−(Σa) and c is in H+(Σb). Consider the non-zero morphismsf : a→ b and g : b→ c. Then the composition gf : a→ c is non-zero.

Proof. (i) This is [19, Lemma 2.5(i)].

(ii) Assume gf is zero. Let h be any morphism in (a, c). By Lemma 3.3.4,h = gf for some morphism f : a→ b. Since (a, b) is one-dimensional,f = αf for some scalar α. Then h = g(αf) = α(gf) would have to bezero. Since h is arbitrary, (a, c) would have to be zero as well, whichis a contradiction since there is c in H−(Σa).

Lemma 3.3.6. Let a, b and c be indecomposable objects of D. Suppose band c are in H+(Σa) and c is in H+(Σb). Consider the non-zero morphism

f : b→ c. Then each morphism a→ c factors as a→ bf→ c.

Proof. This is [19, Lemma 2.5(ii)].

Lemma 3.3.7. Let a, b and c be indecomposable objects of D. Suppose bis in H+(Σa) and c is in both H−(Σa) and H−(Σb). Consider the non-zero

morphism f : b→ c. Then each morphism a→ c factors as a→ bf→ c.

90

Proof. This is to show that (a, f) : (a, b) → (a, c) is surjective. By Serreduality, this is equivalent to (f, Sa) : (c, Sa) → (b, Sa) injective. This map

sends ξ : c → Sa to the composition bf→ c

ξ→ Sa. Since c is in H−(Σa),by Lemma 3.3.3(ii), Sa is in H+(Σc). If ξ is non-zero, then it is a forwardmorphism. Since b is in H+(Σa), by Lemma 3.3.3(i), Sa is in H−(Σb).Therefore by Lemma 3.3.5(ii), the composition ξf is non-zero.

Lemma 3.3.8. (c.f. Lemma 3.2.3) Let a, b and c be indecomposable objectsof D. Suppose b is in H+(Σa) and c is in both H−(Σa) and H−(Σb). Con-sider the non-zero morphisms f : a→ b and g : b→ c. Then the compositiongf : a→ c is non-zero.

Proof. Assume gf is zero. Let h be any morphism in (a, c). By Lemma 3.3.7,h = gf for some morphism f : a→ b. Since (a, b) is one-dimensional, f = αffor some scalar α. Then h = g(αf) = α(gf) would have to be zero. Sinceh is arbitrary, (a, c) would have to be zero as well, which is a contradictionsince there is c in H−(Σa).

However, the composition of non-zero backward morphisms is always zero,and this is stated below. The proof is left to the reader.

Lemma 3.3.9. Let a, b and c be indecomposable objects of D. Suppose bis in H−(Σa) and c is in both H−(Σa) and H−(Σb). Consider the non-zeromorphisms f : a → b and g : b → c. Then the composition gf : a → c iszero.

3.3.1 Lemmas

Consider an infinite band of indecomposable objects, given by (x, y) | y −x ≥ m+ 2, along the top of the Auslander-Reiten quiver of D. Henceforth,let X be add of them.

Lemma 3.3.10. (c.f. Lemma 3.2.4) In the following sketch, the indecom-posable objects of X lie in the region bounded below by and including thedotted boundary line y − x = m + 2. Let a be an indecomposable object inD. Then a has an X -preenvelope a→ u.

Proof. If a = (i, j) is in X , then the lemma is trivial. The following sketchillustrates the situation when a is not in X .

91

w′ w

u

Σ2a a

Consider u = (i, i + m + 2) on the dotted line y − x = m + 2, which isthe leftmost object on the line in the region H+(Σa). Then the non-zeromorphism g : a → u is an X -preenvelope of a. Indeed suppose w is inH+(Σa) ∩ X which is inside H+(Σu). Then each morphism a → w factors

as ag→ u → w (c.f. Lemma 3.3.6). Now suppose w′ is in H−(Σa) ∩ X

which is inside H−(Σu). To show that each morphism a → w′ factors as

ag→ u→ w′ is to show that (g, w′) : (u,w′)→ (a,w′) is surjective. By Serre

duality, this is equivalent to (w′, Sg) : (w′, Sa) → (w′, Su) injective. This

map sends ξ : w′ → Sa to the composition w′ξ→ Sa

Sg→ Su. Since w′ isin H−(Σa), by Lemma 3.3.3(ii), Sa is in H+(Σw′). If ξ is non-zero, then itis a forward morphism. Since g is non-zero, Sg is non-zero. Since u is inH+(Σa), Su is in H+(ΣSa) and thus Sg is also a forward morphism. Finally,

w′ξ→ Sa

Sg→ Su is non-zero by Lemma 3.3.5(i) since Su is in H+(Σw′), andthis holds by Lemma 3.3.3(ii) since w′ is in H−(Σu).

Lemma 3.3.11. (c.f. Lemma 3.2.4) In the following sketch, the indecom-posable objects of X lie in the region bounded below by and including thedotted boundary line y − x = m + 2. Let a be an indecomposable object inD. Then a has an X -precover u→ a.

Proof. If a = (i, j) is in X , then the lemma is trivial. The following sketchillustrates the situation when a is not in X .

w w′

u

a Σ−2a

92

Consider u = (j − m − 2, j) on the dotted line y − x = m + 2, which isthe rightmost object on the line in the region H−(Σ−1a). Then the non-zero morphism g : u → a is an X -precover of a. Indeed suppose w isin H−(Σ−1a) ∩ X which is inside H−(Σ−1u). Then by Lemma 3.3.6 each

morphism w → a factors as w → ug→ a. This is because a is in both H+(Σu)

and H+(Σw), and u is in H+(Σw) (Lemma 3.3.3(ii)). Now suppose w′ is inH+(Σ−1a)∩X which is inside H+(Σ−1u). To show that each morphism w′ →a factors as w → u

g→ a is to show that (w′, g) : (w′, u)→ (w′, a) is surjective.By Serre duality, this is equivalent to (g, Sw′) : (a, Sw′)→ (u, Sw′) injective.

This map sends ξ : a→ Sw′ to the composition ug→ a

ξ→ Sw′. Since w′ is inH+(Σ−1a), Sw′ is in H+(Σa) by Lemma 3.3.3(i), (ii). If ξ is non-zero, then itis a forward morphism. Similarly, g is non-zero, and since u is in H−(Σ−1a),a is in H+(Σu) by Lemma 3.3.3(ii) and thus g is also a forward morphism.

Finally, ug→ a

ξ→ Sw′ is non-zero by Lemma 3.3.5(i) since Sw′ is in H+(Σu),and this holds by Lemma 3.3.3(i), (ii) since w′ is in H+(Σ−1u).

Remark 3.3.12. Alternatively, in Lemma 3.3.11, each morphism w′ → afactors as w → u

g→ a by Lemma 3.3.4, since w′ in H+(Σ−1a) and H+(Σ−1u)implies a, u in H−(Σw′) (Lemma 3.3.3(i)).

By Lemma 3.3.10, Lemma 3.3.11 and [23, Theorem 1.2], the quotient cate-gory DX is pretriangulated. Since τ = Σ and evidently τX = X , thereforeDX is in addition triangulated by [23, Theorem 2.3]. By [23, Theorem 3.2],the Auslander-Reiten quiver of DX is obtained by deleting the vertices ofX and the incident arrows from the Auslander-Reiten quiver of D. Supposethe Auslander-Reiten quiver of DX is of width m vertices. This means DXhas Auslander-Reiten quiver ZAm.

Lemma 3.3.13. (c.f. Lemma 3.2.8) The following sketch shows some in-decomposables in DX .

dm

· · ·

d3

d2

d1

Consider d =

m⊕i=1

di in DX . Then the thick subcategory thick(d) of DX ,

93

generated by d, is in fact equal to DX .

Proof. Similar to Lemma 3.2.8, since DX has Auslander-Reiten triangles by[23, Theorem 3.2].

Lemma 3.3.14. (c.f. Lemma 3.2.9) Consider d =m⊕i=1

di in DX in the

following sketch.

tu

dm

djdi

d1

Then EndDX (d) ∼= EndD(d).

Proof. The idea is that for 1 ≤ i, j ≤ m, there are no non-zero morphismsh : di → dj in D which factor through an object in X . Assume i ≤ j since itis apparent that dj is not in H(Σdi) for j < i. Suppose there is a non-zeroh : di → dj such that h factors through a (not necessarily indecomposable)object t in X as h : di → t→ dj . By Lemma 3.3.10, di has the X -preenvelopedi → u, thus the morphism h further factorizes as h : di → u → t → dj .Since it is again apparent that dj is not in H(Σu), the only morphism fromdi to dj factoring through an object in X is the zero morphism, and theisomorphism exists.

Corollary 3.3.15. (c.f. Corollary 3.2.12) Consider the categories D, X

and DX as usual, and the object d =m⊕i=1

di as in Lemma 3.3.13. Then

EndDX (d) ∼= EndD(d) ∼= (kAm).

Proof. The first isomorphism exists by Lemma 3.3.14. The existence of thesecond isomorphism follows by identifying the di with the pi in Lemma 3.2.11,with the help of Lemma 3.3.1 and Lemma 3.3.5(i).

94


This section displays the significance of Auslander-Reiten triangles in de-termining the mapping cones of certain morphisms in the category D, andthe way Auslander-Reiten triangles alone reflect the (unique) triangulationstructure of the triangulated category D.

The mapping cone constructions realized in this section might not remainthe same in other triangulated categories with Auslander-Reiten quiverswhich contain the same (local) configurations. This is because propertiesof a category, for example, the actions of the translation functor and theAuslander-Reiten translation, are inherent attributes of the quiver.

Consider the Auslander-Reiten quiver of D in the following diagram.

...

•

•

•

•

•

•

??

•

??

•

??

•

??

•

??

•

•

??

•b2

??

•

??

•

??

•

??

. . . •

??

•a2

g2

f2 ??

•b1

??

•

??

•

??

• . . .

•

??

•a1

g1

f1 ??

•b0

??

•

??

•

??

•

??

•

??

•a0

f0 ??

•c??

•

??

•

The coordinates of some of the objects are as follows: a0 = (i, j), b0 =(i, j + 1), a1 = (i− 1, j), b1 = (i− 1, j + 1), a2 = (i− 2, j), b2 = (i− 2, j + 1)and c = (i + 1, j + 1). As usual, the coordinates of objects on the bottomline satisfy the equation y − x = 2.

Lemma 3.3.16. The mapping cones of the maps fn : (i−n, j)→ (i−n, j+1)are all isomorphic to c = (i+ 1, j + 1).

Proof. Since a0 → b0 → c→ is an Auslander-Reiten triangle, the statementis true for n = 0. Suppose the statement is true for n = p, p ≥ 0, i.e. themapping cone of the map fp : ap → bp is c = (i+ 1, j + 1). To see that thestatement is true for n = p+1, consider the following commutative diagram,

ap+1µp+1 // ap ⊕ bp+1

ap+1

fp+1

// bp+1,

95

where µp+1 =

(gp+1

fp+1

)and the map is the canonical surjection. By the

octahedral axiom, it may be extended to the following commutative diagram,

0 //

ap

ı

ap

fp

// 0

ap+1

µp+1 // ap ⊕ bp+1//

bp

// Σap+1

ap+1fp+1

//

bp+1

// c

// Σap+1

0 // Σap Σap // 0,

where the map ı is the canonical injection. The distinguished triangle onthe second row is an Auslander-Reiten triangle. Hence the mapping cone ofthe map fp+1 : ap+1 → bp+1 is the same as the mapping cone of the mapfp : ap → bp, which is c = (i+ 1, j + 1) by the induction hypothesis.

Remark 3.3.17. The connecting morphism ∂ in the Auslander-Reiten trian-

gle ap+1µp+1→ ap ⊕ bp+1 → bp

∂→ Σap+1 on the second row is a backwardmorphism.

Lemma 3.3.18. Consider the following sketch.

•(i,j+1)

•(i,j−r)

??

•cr

•c0 y−x=2

For −1 ≤ n ≤ j − i − 2, let an = (i, j − n). Let fn be the morphismfn : (i, j − n)→ (i, j − n+ 1). Then for 0 ≤ r ≤ j − i− 2, the compositionf0 . . . fr−1fr : (i, j − r)→ (i, j + 1) has mapping cone cr = (j − 1− r, j + 1).

Proof. Consider the object c0 = (j − 1, j + 1) on the bottom line y − x =2. It lies in the same descending line as (i, j + 1), where the composition

96

f0 . . . fr−1fr maps to. The mapping cone cr = (j − 1 − r, j + 1) lies in thesame descending line as c0, but it is r steps up the line. This is how weunderstand the location of the mapping cone cr.

Let us consider again the Auslander-Reiten quiver of D.

...

•

•

•

•

•

•

•

??

•Σa−1

??

•a−1

??

•

??

•

??

. . . •

??

•Σa0

??

•a0

f0 ??

•

??

•

??

• . . .

•Σa1

??

•a1

f1 ??

•

??

•

??

•

??

•

??

•

??

•

??

•

??

•

??

•

•

??

•

??

•

??

•

??

•c1

??

•

??

•

??

•

??

•

??

•d0

??

•c0

The statement is true for r = 0 by Lemma 3.3.16. Suppose the statement istrue for r = p, p ≥ 0, and then the statement is also required to be true forr = p+ 1.

Consider the following commutative diagram,

ap+1g // a0

f0

ap+1

f0g// a−1,

where g is the morphism f1 . . . fp+1 : (i, j − p− 1)→ (i, j).

By the octahedral axiom, it may be extended to the following commutativediagram,

ap+1g // a0

f0

ρ // d

// Σap+1

ap+1f0g //

a−1//

∗

// Σap+1

0 //

c0

%

c0

(Σρ)%

// 0

Σap+1

// Σa0Σρ // Σd // Σ2ap+1.

97

Consider the object d0 = (j − 2, j) on the bottom line y − x = 2. It liesin the same descending line as (i, j), where the morphism g maps to. Bythe induction hypothesis, the mapping cone of the morphism g lies in thesame descending line as d0, but it is p steps up the line, i.e. it is the objectd = (j−2−p, j). This gives the distinguished triangle on the first row. Themapping cone of the morphism f0 is c0 = (j − 1, j + 1) by Lemma 3.3.16,which gives the distinguished triangle in the second column.

The map ρ : a0 → d is non-zero, as otherwise ap+1∼= Σ−1d ⊕ a0 by

Lemma 0.2.2(v), which is not possible since ap+1 is indecomposable. There-fore the map Σρ : Σa0 → Σd is non-zero as well. Similarly, the map% : c0 → Σa0 is non-zero. Therefore by Lemma 3.3.5(ii) the composition(Σρ)% : c0 → Σd is non-zero, and the distinguished triangle d → ∗ → c0 →Σd is non-split.

Let an object e have coordinates (j − 2 − p, j + 1). By Lemma 3.3.16, themapping cone of d → e is c0. Since (c0,Σd) is one-dimensional, the object∗ is indeed equal to e. Therefore the mapping cone of the morphism f0g ise = (j − 2− p, j + 1) = (j − 1− (p+ 1), j + 1) = cp+1 as desired.

Using the same notation as in [23, Theorem 2.3], let σ be the translationfunctor of the triangulated category DX . The construction of the functor σis described in [23, Setup 1.1]. The following example helps understand it.

Example 3.3.19. Consider again the Auslander-Reiten quiver in Lemma 3.3.18with the same coordinate system.

...

•

•

•

•

•

•

•

??

•

??

•a−1

??

•

??

•

??

. . . •

??

•

??

•a0

f0 ??

•

??

•

??

• . . .

•Σa1

??

•a1

f1 ??

•

??

•c3

??

•

??

•Σa2

??

•a2

f2 ??

•

??

•d2

??

•c2

??

•

•a3

f3 ??

•

??

•

??

•d1

??

•c1

??

•

??

•

??

•

??

•

??

•d0

??

•c0

Consider an infinite band of indecomposable objects along the top of theAuslander-Reiten quiver bounded below by and including the dotted bound-

98

ary line, i.e. the region given by (x, y) | y − x ≥ j + 1 − i, and let X beadd of them.

For example, the morphism f1 : a1 → a0 is an X -monomorphism in D, i.e.a morphism such that each morphism a1 → x with x in X factors throughf1 (c.f. Lemma 3.3.10). Extend f1 : a1 → a0 to the distinguished triangle

a1f1→ a0 → d0

∂1→ Σa1 in D by Lemma 3.3.18. On the other hand, themapping cone of the morphism f0f1 is c1 by Lemma 3.3.18, where f0f1 :a1 → a−1 is an X -preenvelope of a1 by Lemma 3.3.10. Then the diagram

a1f1→ a0 → d0

∂2→ c1, considered in DX , is defined to be a distinguishedtriangle in DX , so that σ(a1) = c1. The reader can refer to [23, Setup 1.1]for more details.

The following deserves attention.

(i) The connecting morphism ∂1 in the distinguished triangle a1f1→ a0 →

d0∂1→ Σa1 in D is a backward morphism, while the connecting mor-

phism ∂2 in the distinguished triangle a1f1→ a0 → d0

∂2→ c1 in DX is aforward morphism.

(ii) The morphism f0f1 : a1 → a−1 is an X -preenvelope of a1, hence an

X -monomorphism as well. Therefore a1f0f1→ a−1 → c1

∂→ c1 is adistinguished triangle in DX by construction. However, a−1 in X isisomorphic to zero in DX , so that the connecting morphism ∂ has tobe an isomorphism in DX .

(iii) Similarly, to suggest but a few, there is the distinguished triangle

a2f1f2→ a0 → d1 → c2 in DX , and also a3

f1f2f3→ a0 → d2 → c3 inDX . The distinguished triangles in DX seem to be quite symmetrical.

Lemma 3.3.20. (c.f. Lemma 3.2.13) For i+ 2 ≤ j′ ≤ i+m+ 1, we haveσ(i, j′) = (j′ − 1, i+m+ 2).

Proof. By Lemma 3.3.10, (i, j′)→ (i, i+m+2) is an X -preenvelope. Extendthe map (i, j′)→ (i, i+m+ 2) to the distinguished triangle (i, j′)→ (i, i+m+ 2)→ (j′ − 1, i+m+ 2)→ given in Lemma 3.3.18, which is in the formof the distinguished triangle (2) in [23, Setup 1.1] when M is replaced by(i, j′). Hence σ(i, j′) = (j′ − 1, i+m+ 2).

Remark 3.3.21. The mapping cone construction described in this sectioncan be imitated to accommodate the previous situation in Section 3.2.2.An example in the finite derived category Db(mod kA7) is given in the Ap-pendix (Lemma A.3.1 and Lemma A.3.2). Since there is a high resemblance

99

between the respective locations of the X -preenvelopes (Lemma 3.2.4 andLemma 3.3.10), there is no surprise that Lemma 3.2.13 and Lemma 3.3.20yield the same consequence accordingly.

Corollary 3.3.22. (c.f. Corollary 3.2.15) For i + 2 ≤ j′ ≤ i + m + 1 andN ≥ 0 an integer,

σN (i, j′) =

(i+ N

2 (m+ 1), j′ + N2 (m+ 1)) for N even,

(j′ + N−12 (m+ 1)− 1, i+ N+1

2 (m+ 1) + 1) for N odd.


For example, for i+2 ≤ j′ ≤ i+m+1, we have σ2(i, j′) = (i+m+1, j′+m+1).


Lemma 3.3.23. (c.f. Lemma 3.2.16) For j −m− 1 ≤ i′ ≤ j − 2, we haveσ−1(i′, j) = (j −m− 2, i′ + 1).


Lemma 3.3.23 is deduced from Lemma 3.3.20, where X is perceived as preen-veloping. Alternatively, one can obtain the value of σ−1 directly if X isperceived as precovering. An example is given in the Appendix (Exam-ple A.3.8).

Corollary 3.3.24. (c.f. Corollary 3.2.18) For j −m − 1 ≤ i′ ≤ j − 2 andN ≥ 0 an integer,

σ−N (i′, j) =

(i′ − N

2 (m+ 1), j − N2 (m+ 1)) for N even,

(j − N+12 (m+ 1)− 1, i′ − N−1

2 (m+ 1) + 1) for N odd.


Lemma 3.3.25. (c.f. Lemma 3.2.19) Let N ≥ 1 be an integer. Consider

d =i+m+1⊕j′=i+2

(i, j′) in DX . Then HomDX (d, σNd) = 0.

Proof. This is similar to Lemma 3.2.19.

Lemma 3.3.26. (c.f. Lemma 3.2.20) Let N ≥ 1 be an integer. Consider

d =

i+m+1⊕j′=i+2

(i, j′) in DX . Then HomDX (d, σ−Nd) = 0.

100

Proof. This is similar to Lemma 3.2.20. However, for i + 2 ≤ j′ ≤ i + m +1, any non-zero morphism in D from (i, j′) to an object y in H−(Σ(i, j′))has to factor through an object in X . This is true by Lemma 3.3.4, sincethere is always a non-zero forward morphism u : x → y where x is anindecomposable object in X and in H−(Σ(i, j′)). An example is given in thefollowing diagram, where the indecomposable objects of X lie in the regionbounded below by and including the dotted boundary line.

x

y

Σ2(i, j′) (i, j′)

By Lemma 3.3.1, the non-zero morphism space from (i, j′) to y is one-dimensional. Therefore there are no non-zero morphisms in DX from (i, j′)to an object y in H−(Σ(i, j′)).

3.3.3 Theorem


Theorem 3.3.27. Consider the categories D, X and DX as usual, and the

object d =m⊕i=1

di as in Lemma 3.3.13. Then there is an equivalence of

triangulated categories f : DX'→Db(mod (kAm)) where f(d) = (kAm).

Proof. Let U and X in Theorem 3.1.1 be DX and d respectively. By Corol-lary 3.3.15, A ∼= (kAm). Let us take Σ in Theorem 3.1.1 to be the trans-lation functor σ of the triangulated category DX . Then the result followsfrom Lemma 3.3.13, Lemma 3.3.25 and Lemma 3.3.26.

3.4 The finite derived category Db(mod kD5)

In this section, let T be the finite derived category Db(mod kD5) describedin Section 1.4. The category T is a k-linear Hom finite triangulated category.It is Krull-Schmidt as well. The Auslander-Reiten quiver of Db(mod kD5)is given below (Section 1.4).

101

•m+

•a+

α′+

•b+

•c+

•d+

•e+

•m−

•a−

α′−

•b−

•c−

•d−

•e−

. . . •m3

??

GG

•a3

α′2

α−

??α+

GG

•b3

??

GG

•c3

??

GG

•d3

??

GG

•e3??

GG

•f3

. . .

•m2

??

•a2

α′1

α2 ??

•b2

??

•c2

??

•d2

??

•e2??

•f2

??

•

•a1

α1 ??

•b1??

•c1??

•d1

??

•e1??

•f1

??

•g1

??

Let i ∈ 1, 2, 3 and j ∈ 2, 3,+,−. If there is an arrow bi → bj , then letthe arrow be βi : bi → bj if j ∈ 2, 3 and let the arrow be βj : bi → bj ifj ∈ +,−. Now let i ∈ 2, 3,+,− and j ∈ 1, 2, 3. If there is an arrowbi → cj , then let the arrow be β′i−1 : bi → cj if i ∈ 2, 3 and let the arrowbe β′i : bi → cj if i ∈ +,−. The rest of the arrows are named similarlyin terms of γ’s, δ’s, etc. The vertices m+,m−, a+, a−, b+, b−, . . . are theexceptional vertices. The vertices a3, b3, c3, . . . are the bridge vertices. Theexceptional vertices together with the bridge vertices are said to lie in theexceptional part of the Auslander-Reiten quiver. The vertices other than theexceptional vertices are the type A vertices, and are said to lie in the typeA part of the Auslander-Reiten quiver. The exceptional and type A partsof the Auslander-Reiten quiver are not disjoint. The vertices a3 and b3 areassigned to be one horizontal unit apart, and so are a+ and b+ etc.

The following coordinate system is employed. Suppose the indecomposableobject a3 has coordinates (i − 1, i + 3). The coordinates of some of itssurrounding objects are given in the following diagram.

(i− 2, i+ 3)+

(i− 1, i+ 4)+

(i, i+ 5)+

(i− 2, i+ 3)−

(i− 1, i+ 4)−

(i, i+ 5)−

(i− 2, i+ 2)

??

GG

(i− 1, i+ 3)

??

GG

(i, i+ 4)

??

GG

(i+ 1, i+ 5)

(i− 1, i+ 2)

??

(i, i+ 3)

??

(i+ 1, i+ 4)

??

(i− 1, i+ 1)

??

(i, i+ 2)

??

(i+ 1, i+ 3)

??

(i+ 2, i+ 4)

The coordinates of objects of the bottom line satisfy the equation y−x = 2,and the coordinates of the exceptional vertices satisfy the equation y−x = 5.

The action of the translation functor Σ on the Auslander-Reiten quiver is

102

given in [33, Table 4.I.]. It acts by shifting 4 units to the right and switchingeach pair of exceptional vertices. For example, Σ(b2) = f2 and Σ(a+) = e−.

The following is an example of Lemma 1.4.1 and the narration after it, andis rewritten below to remind the reader.

Lemma 3.4.1. (c.f. Lemma 3.2.1) Let x and y be indecomposable objectsof Db(mod kD5). Then by [14, 4.6], any non-zero morphism f from x to yis a linear combination of morphisms, written f = Σαifi, where the αi arescalars and the fi : x → y are compositions of irreducible morphisms. Thefinite derived category Db(mod kD5) is also standard.

Definition 3.4.2. Let a be an indecomposable object of Db(mod kD5), andlet L(a) be the set of indecomposable objects with non-zero morphisms toa. Dually, let R(a) be the set of indecomposable objects to which there arenon-zero morphisms from a.

The Auslander-Reiten quiver contains different types of local configurationswhich give rise to different Auslander-Reiten triangles and commutativityrelations. In the configurations below, the black dots indicate the bridgevertices.

x

v′′

x′

y

v′

y′

(i)

•au′

??u′′

GG

u

•b

w′′

GG

w′

??

w

•c

zv

??

z′

??

In Configuration (i), since a → x ⊕ y ⊕ z → b → is an Auslander-Reitentriangle, v′′u′′+v′u′+vu = 0. There are also the Auslander-Reiten trianglesx→ b→ x′ → and y → b→ y′ →, and so w′′v′′ = 0 and w′v′ = 0.

•bt

•b′t′

a

s??

q

ds′??

q′

e (ii)

cr

??

c′r′

??

In Configuration (ii), since a→ b⊕c→ d→ is an Auslander-Reiten triangle,ts+ rq = 0. There is also the Auslander-Reiten triangle c→ d→ c′ →, andso q′r = 0.

103

The above description permits us to give examples of the region R(x) forsome indecomposable objects x in Db(mod kD5). The way they are derivedis very similar to the situation given in Example 1.4.4.

Diagram 3.4.3. The region R(a3) is as shown by the black dots in thefollowing diagram.

m+

•a+

•b+

•c+

d+

e+

m−

•a−

•b−

•c−

d−

e−

. . . m3

??

GG

•a3

??

GG

•b3

??

GG

•c3

??

GG

•d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

•b2

??

•c2

??

•d2

??

e2??

f2

??

a1

??

b1??

•c1??

•d1

??

e1??

f1

??

g1

??

Diagram 3.4.4. The region R(a−) is as shown by the black dots in thefollowing diagram.

m+

a+

•b+

c+

•d+

e+

m−

•a−

b−

•c−

d−

e−

. . . m3

??

GG

a3

??

GG

•b3

??

GG

•c3

??

GG

•d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

b2

??

•c2

??

•d2

??

e2??

f2

??

a1

??

b1??

c1??

•d1

??

e1??

f1

??

g1

??

Diagram 3.4.5. The region R(a1) is as shown by the black dots in thefollowing diagram.

m+

•a+

b+

c+

d+

e+

m−

•a−

b−

c−

d−

e−

. . . m3

??

GG

•a3

??

GG

•b3

??

GG

c3

??

GG

d3

??

GG

e3??

GG

f3

. . .

m2

??

•a2

??

b2

??

•c2

??

d2

??

e2??

f2

??

•a1

??

b1??

c1??

•d1

??

e1??

f1

??

g1

??

With the following lemma we end this section.

104

Lemma 3.4.6. (c.f. Lemma 3.2.3) Let a, b and c be indecomposable objectsof Db(mod kD5). Assume (a, b), (b, c) and (a, c) are all one-dimensional.Let f : a → b and g : b → c be some non-zero morphisms. Then thecomposition gf : a→ c is non-zero.


3.4.1 Hom spaces

In this section, the non-zero morphisms between certain indecomposableobjects are described. This will lead to the calculations of the dimensionsof some of the Hom spaces.

Lemma 3.4.7. Suppose a is a bridge vertex and b is an exceptional vertexwhere b is in R(a). Then any non-zero morphism f : a → b can be writtenas a linear combination of paths lying in the exceptional part of the quiver.

Proof. Let y be the bridge vertex with an arrow to b. Any non-zero mor-phism f : a → b can be factored as a

g→ y → b by Lemma 3.4.1, since y isthe only vertex with an arrow to b. Consider g : a → y. Write g = Σαipi,where αi is a scalar and pi is a path from a to y. It is enough to show thatthe path pi is a linear combination of paths lying in the exceptional part ofthe quiver.

If in the path pi there are consecutive arrows q and r placed as follows,where the black dot is a bridge vertex,

•t

s ??

q

,

r

??

then rq can be replaced by −ts, by virtue of the little remark after Con-figuration (ii). And if in the path pi there are consecutive arrows u and vplaced as follows, where the black dots are the bridge vertices,

v′′

v′ • u′

??u′′

GG

u

• ,

v

??

then vu can be replaced by −(v′u′ + v′′u′′), by virtue of the little remarkafter Configuration (i). In this way, the path pi is transformed into a linearcombination of paths lying in the exceptional part of the quiver.

105

Lemma 3.4.8. Suppose a is an exceptional vertex and b is a type A vertexwhere b is in R(a). Then any non-zero morphism f : a → b can be writtenf = Σαifig, where αi is a scalar, g is an irreducible morphism from thevertex a to a bridge vertex y and fi : y → b is a path lying in the type A partof the quiver.

Proof. It is sufficient to consider the example a = a−. The region R(a−) isgiven in Diagram 3.4.4, which is as shown by the black dots in the followingdiagram.

m+

a+

•b+

c+

•d+

e+

m−

•a−

b−

•c−

d−

e−

. . . m3

??

GG

a3

??

GG

•b3

??

GG

•c3

??

GG

•d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

b2

??

•c2

??

•d2

??

e2??

f2

??

a1

??

b1??

c1??

•d1

??

e1??

f1

??

g1

??

The bridge vertex y is b3 and g = α′− here. The lemma is immediate whenb = b3, c2 and d1. Suppose b = c3, then write f = Σαipiα

′−, where pi is

a path from b3 to c3. Then it remains to show that piα′− can be written

piα′− = fiα

′−, where fi is a path lying in the type A part of the quiver.

However, pi can only be

(i) pi = β′+β+,

(ii) pi = β′−β−,

(iii) pi = γ2β′2.

For (iii), the path pi already lies in the type A part of the quiver. For (ii),we have however piα

′− = β′−β−α

′− = 0 by virtue of the little remark after

Configuration (i). Finally, for (i), we have piα′− = β′+β+α

′− = −(β′−β− +

γ2β′2)α′− = −γ2β

′2α′− by virtue of the little remark after Configuration

(i), and γ2β′2 lies in the type A part of the quiver. The case where b = d2

or b = d3 is similar.

Corollary 3.4.9. Suppose a is a bridge vertex and b is an exceptional vertexwhere b is in R(a). Then (a, b) is one-dimensional.

Proof. It is sufficient to consider the example a = a3. The region R(a3) isgiven in Diagram 3.4.3, which is as shown by the black dots in the followingdiagram.

106

m+

•a+

•b+

•c+

d+

e+

m−

•a−

•b−

•c−

d−

e−

. . . m3

??

GG

•a3

??

GG

•b3

??

GG

•c3

??

GG

•d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

•b2

??

•c2

??

•d2

??

e2??

f2

??

a1

??

b1??

•c1??

•d1

??

e1??

f1

??

g1

??

If b = a+ or b = a−, then it is true by Lemma 3.4.1. Consider a non-zero morphism f : a → b. Then by Lemma 3.4.7, assume f to be a linearcombination of paths lying in the exceptional part of the quiver. If b =b+, then there is only one non-zero path β+α

′−α− : a → b lying in the

exceptional part of the quiver. This is because the other path β+α′+α+ :

a → b is a zero path by virtue of the little remark after Configuration (i).The case where b = b−, b = c+ or b = c− is similar.

Corollary 3.4.10. Suppose a is an exceptional vertex and b is a type Avertex where b is in R(a). Then (a, b) is one-dimensional.

Proof. Consider a non-zero morphism f : a → b. Then by Lemma 3.4.8, fcan be written f = Σαifig, where αi is a scalar, g is an irreducible morphismfrom a to a bridge vertex y and fi : y → b is a path lying in the type Apart of the quiver. And the result is immediate from the little remark afterConfiguration (ii).

Corollary 3.4.11. Let a and b be two exceptional vertices where b is inR(a). Then (a, b) is one-dimensional.

Proof. Consider a non-zero morphism f : a → b. Then f can be writtenf = Σαifif

′, where αi is a scalar, f ′ : a→ y is a morphism from the vertexa to a bridge vertex y, because there is only one arrow going from a. Sincefi is then a morphism from the bridge vertex y to the exceptional vertex b,the rest is very similar to Corollary 3.4.9.

Corollary 3.4.12. The Hom space (a1, d1) is one-dimensional.

Proof. The region R(a1) is given in Diagram 3.4.5. Consider a non-zeromorphism f : a1 → d1. Then write f = Σαipi as a linear combination ofpaths with pi : a1 → d1. The path pi cannot contain the vertices b1, b2 norc1, since they are not in R(a1), therefore it has to either go through thevertex a+ or the vertex a−. However, the two choices are equal up to signs.

107

For example, consider a path pi which goes through the vertex a+. Sincethe vertex b2 is not in R(a1), it can be written as a path which goes throughthe vertex a− by virtue of the little remark after Configuration (i).

3.4.2 Lemmas

Consider the indecomposable objects a+, b+, c+, . . . along the top line of theAuslander-Reiten quiver of T . Henceforth, let X be add of them.

Lemma 3.4.13. The indecomposable object a3 has an X -preenvelope a3 →a+ ⊕ b+.

Proof. The objects in X to which there is a non-zero morphism from a3

are a+, b+ and c+. The cases for a+ and b+ are trivial. Suppose a non-zero morphism h : a3 → c+ is given. It is enough to assume that h is acomposition of irreducible morphisms and to show that h factors througha3 → a+. This is true by Lemma 3.4.7 and Corollary 3.4.9.

Lemma 3.4.14. The indecomposable object a2 has an X -preenvelope a2 →a+ ⊕ b+.

Proof. This is because the intersection of R(a2) and X only consists ofa+ and b+, and that (a+, b+) is zero by virtue of the little remark afterConfiguration (i).

Lemma 3.4.15. The indecomposable object a− has an X -preenvelope a− →b+.

Proof. This is because the intersection of R(a−) and X is the same as theintersection of R(b+) and X (the intersection consists of b+ and d+), andsince (a−, d+) is one-dimensional by Corollary 3.4.11, any non-zero mor-phism f : a− → d+ has to factor through b+. This is in the proof ofCorollary 3.4.11 as well.

Lemma 3.4.16. The indecomposable object a1 has an X -preenvelope a1 →a+.


By Lemma 3.4.13, Lemma 3.4.14, Lemma 3.4.15, Lemma 3.4.16 and [23,Theorem 1.2], the quotient category TX is pretriangulated. Since evidentlyτX = X , therefore TX is in addition triangulated by [23, Theorem 2.3]. By[23, Theorem 3.2], the Auslander-Reiten quiver of TX is obtained by deleting

108

vertices on the top line and the incident arrows from the Auslander-Reitenquiver of T . Therefore the two categories TX and Db(mod kA4) have thesame Auslander-Reiten quiver ZA4.

Definition 3.4.17. Let a be an indecomposable object in TX . Similarly toDefinition 3.4.2, let LX (a) be the set of indecomposable objects with non-zero morphisms to a in TX . Dually, let RX (a) be the set of indecomposableobjects to which there are non-zero morphisms from a in TX .

Lemma 3.4.18. (c.f. Lemma 3.2.7) The region RX (a3) is as shown by theblack dots in the following diagram.

m+

a+

b+

c+

d+

e+

m−

•a−

b−

c−

d−

e−

. . . m3

??

GG

•a3

??

GG

•b3

??

GG

c3

??

GG

d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

•b2

??

•c2

??

d2

??

e2??

f2

??

a1

??

b1??

•c1??

d1

??

e1??

f1

??

g1

??

Proof. The philosophy of Section 3.4.1 allows us to see that the regionRX (a3) is at most the region as indicated. Now let s be an object in it.Since s is in R(a3), there is a non-zero morphism f : a3 → s in T so thatthe corresponding morphism f : a3 → s is non-zero in TX . Suppose other-wise that f is zero. Then f is a morphism in T which factors through a (notnecessarily indecomposable) object t in X , i.e. f factorizes as f : a3 → t→ sin T . By Lemma 3.4.13, a3 has the X -preenvelope a3 → a+ ⊕ b+, thereforef further factorizes as f : a3 → a+ ⊕ b+ → t → s in T . The morphisma+ ⊕ b+ → t is non-zero, therefore either a+ → t is non-zero, or b+ → tis non-zero. In the first case, t is to be (finite sums of) c+, but s is not inR(c+). In the second case, t is to be (finite sums of) d+, but s is not inR(d+) either. This gives f non-zero in TX . Also Lemma 3.4.6 is implicitlyused throughout.


m+

a+

b+

c+

d+

e+

m−

•a−

b−

c−

d−

e−

. . . m3

??

GG

•a3

??

GG

•b3

??

GG

c3

??

GG

d3

??

GG

e3??

GG

f3

. . .

m2

??

•a2

??

•b2

??

c2

??

d2

??

e2??

f2

??

a1

??

•b1??

c1??

d1

??

e1??

f1

??

g1

??

109

Proof. Similar.


m+

a+

b+

c+

d+

e+

m−

•a−

b−

c−

d−

e−

. . . m3

??

GG

•a3

??

GG

b3

??

GG

c3

??

GG

d3

??

GG

e3??

GG

f3

. . .

m2

??

•a2

??

b2

??

c2

??

d2

??

e2??

f2

??

•a1

??

b1??

c1??

d1

??

e1??

f1

??

g1

??

Proof. Similar.

Lemma 3.4.21. (c.f. Lemma 3.2.7) The region RX (a−) is as shown by theblack dots in the following diagram.

m+

a+

b+

c+

d+

e+

m−

•a−

b−

c−

d−

e−

. . . m3

??

GG

a3

??

GG

•b3

??

GG

c3

??

GG

d3

??

GG

e3??

GG

f3

. . .

m2

??

a2

??

b2

??

•c2

??

d2

??

e2??

f2

??

a1

??

b1??

c1??

•d1

??

e1??

f1

??

g1

??

Proof. Similar.

Lemma 3.4.22. (c.f. Lemma 3.2.8) Consider a = a1⊕a2⊕a3⊕a− in TX .Then the thick subcategory thick(a) of TX , generated by a, is in fact equal toTX .

Proof. Similar to Lemma 3.2.8, since TX has Auslander-Reiten triangles by[23, Theorem 3.2].

Lemma 3.4.23. (c.f. Lemma 3.2.9) Consider a = a1⊕a2⊕a3⊕a− in TX .Then EndTX (a) ∼= EndT (a).

Proof. This is similar to Lemma 3.2.9. But surely, we shall do it again.The idea is that there are no non-zero morphisms h : ai → aj in T , wherei, j ∈ 1, 2, 3,−, which factor through an object in X . Only the cases wherei, j ∈ 1, 2, 3 or j = − need to be considered, since there are no non-zeromorphisms from ai to aj otherwise. Assume i ≤ j if i, j ∈ 1, 2, 3. Suppose

110

there is a non-zero h : ai → aj such that h factors through an object y in Xas h : ai → y → aj . By Lemma 3.4.13, Lemma 3.4.14, Lemma 3.4.15 andLemma 3.4.16, ai has an X -preenvelope ai → x, thus the morphism h furtherfactorizes as h : ai → x → y → aj . Since x can only be a+, b+ or a+ ⊕ b+,there are apparently no non-zero morphisms from x to aj . Therefore theonly morphism from ai to aj factoring through an object in X is the zeromorphism, and the isomorphism exists.

Corollary 3.4.24. (c.f. Corollary 3.2.12) Consider the categories T , X andTX as usual, and the object a = a1⊕a2⊕a3⊕a− in TX as in Lemma 3.4.22.Then EndTX (a) ∼= EndT (a) ∼= (kA4).

Proof. The first isomorphism exists by Lemma 3.4.23. The existence ofthe second isomorphism follows by identifying the ai with the pi for 1 ≤i ≤ 3 and by identifying a− with p4 in Lemma 3.2.11, with the help ofLemma 3.4.6.


Lemma 3.4.25. The mapping cone of the morphism β′1 : b2 → c1 is Σb1.

Proof. This is because b1 → b2 → c1 → is an Auslander-Reiten triangle.

Lemma 3.4.26. The mapping cone of the morphism β′2 : b3 → c2 is Σb1.

Proof. By the octahedral axiom, extend to the following commutative dia-gram,

b1 //

b3

β′2 // c2// ∗

b2

// b3 ⊕ c1//

c2

//

c1

c1// 0

Σb1 .

The distinguished triangle on the second row is an Auslander-Reiten trian-gle, and the distinguished triangle on the first column is given by Lemma 3.4.25.

Lemma 3.4.27. The mapping cone of the morphism γ′1β′2 : b3 → d1 is Σb2.

111


b1

∗

// b3γ′1β′2 //

d1// Σ∗

c1

// c2//

d1

// Σc1

Σb1 Σb1 // 0 .

The distinguished triangle on the third row is given by Lemma 3.4.25 andthe distinguished triangle in the second column is given by Lemma 3.4.26.

By Corollary 3.4.12, (c1,Σb1) = (c1, f1) is one-dimensional (replacing a1 byc1 and d1 by f1). Since b3 is indecomposable, the distinguished triangleb1 → b3 → c2 → Σb1 is non-split, and the morphism c2 → Σb1 is non-zero.Similarly, since d1 is indecomposable, the morphism c1 → c2 is non-zero,hence the morphism c1 → Σb1 is non-zero by Lemma 3.4.6. Therefore thedistinguished triangle in the first column has to be the Auslander-Reitentriangle b1 → b2 → c1 →.

Lemma 3.4.28. The mapping cone of the morphism µ : a+ → d1 is d+.

Proof. The Hom space (a+, d1) is one-dimensional by Corollary 3.4.10, there-fore the mapping cones of all non-zero morphisms µ : a+ → d1 are iso-morphic. Such considerations permeate this section, and they will not berepeated each time, since most Hom spaces are one-dimensional.

In the following diagram, the region R(d1) is as shown by the black dots.

m+

a+

b+

c+

•d+

e+

m−

a−

b−

c−

•d−

e−

. . .

??

GG

a3

??

GG

b3

??

GG

c3

??

GG

•d3

??

GG

•e3??

GG

f3

. . .

??

a2

??

b2

??

c2

??

•d2

??

e2??

•f2

??

a1

??

b1??

c1??

•d1

??

e1??

f1

??

•g1

??

Extend a non-zero morphism µ to the distinguished triangle a+µ→ d1

ν→∗ →. By [16, Lemma 6.5], the mapping cone ∗ is indecomposable. The

112

morphism ν is non-zero, otherwise the distinguished triangle Σ−1∗ → a+µ→

d1ν→ ∗ splits, but this is impossible since a+ is indecomposable. Therefore

the object ∗ has to be in R(d1).

The object ∗ cannot be d1, d2, d3 or d−. Assume it is. Since (a+, d1) is one-dimensional by Corollary 3.4.10, (d1, ∗) is one-dimensional by Lemma 3.4.1and (a+, ∗) is one-dimensional by Corollary 3.4.10 and Corollary 3.4.11, thecomposition νµ is non-zero by Lemma 3.4.6. However, this is a contradictiongiven that the composition of two consecutive morphisms of a distinguishedtriangle is zero by Lemma 0.2.2(i).

Appealing to the octahedral axiom, extend to the following commutativediagram,

Σ−1∗ //

a+

µ // d1ν // ∗

b2 //

b3 //

d1

// Σb2

b+

b+

// 0

∗ .

The distinguished triangle on the second row is given in Lemma 3.4.27. Thedistinguished triangle in the second column is an Auslander-Reiten triangle.Suppose the object ∗ is e3. Then the distinguished triangle on the first rowis a3 → a+

µ→ d1ν→ e3. But again this is not possible for the same reason.

Consider the distinguished triangle in the first column. Since b2 is indecom-posable, the distinguished triangle is non-split. This means the morphismfrom b+ to ∗ is non-zero. The object ∗ cannot be f2 or g1, since neither ofthem is in R(b+). Hence the mapping cone has to be the object d+.

Corollary 3.4.29. The mapping cone of the morphism µ : b2 → b+ is d+.

Proof. Consider the distinguished triangle Σ−1∗ → b2 → b+ → ∗ in thefirst column of the commutative diagram in Lemma 3.4.28, and the resultis evident. Also the Hom space (b2, b+) is one-dimensional.

Lemma 3.4.30. The mapping cone of the morphism µ : m+ → b2 is b−.

Proof. The Hom space (m+, b2) is one-dimensional by Corollary 3.4.10. By

113

the octahedral axiom, extend to the following commutative diagram,

m+

µ // b2

// ∗

0 // c1

c1

// 0

∗ // c+

//

Σb1

// Σ∗

.

The distinguished triangle m+ → c1 → c+ → is given by Lemma 3.4.28 andthe distinguished triangle b2 → c1 → Σb1 → is given by Lemma 3.4.25. Fi-nally, the mapping cone of the morphism c+ → Σb1 is Σb− by Lemma 3.4.28.Also the Hom space (c+,Σb1) is one-dimensional.

Lemma 3.4.31. The mapping cone of the morphism α+ : a3 → a+ is d−.

Proof. By the octahedral axiom, extend to the following commutative dia-grams,

0 //

a− ⊕ b2

a− ⊕ b2

//

a3// a+ ⊕ a− ⊕ b2 //

b3

//

a3α+ // a+

//

∗

//

,

// b2

b2

// 0

// a− ⊕ b2 //

b3 //

∗

// a− //

d1

// ∗

.

In the first diagram, the distinguished triangle on the second row is anAuslander-Reiten triangle. In the second diagram, the distinguished triangle

114

in the middle column is given by Lemma 3.4.27. Finally, the mapping coneof the morphism a− → d1 is d−, by a mirror version of Lemma 3.4.28.

Corollary 3.4.32. The mapping cone of the morphism a2 → a+ is c+.

Proof. It is the same as Corollary 3.4.29 by a suitable translation of coordi-nates.

Corollary 3.4.33. The mapping cone of the morphism a1 → a+ is b−.

Proof. The Hom space (a1, a+) is one-dimensional by Lemma 3.4.1. By theoctahedral axiom, extend to the following commutative diagram,

a1

// a+// ∗ //

Σ−1c+//

a2//

a+

// c+

b1

b1 // 0

∗ .

The distinguished triangle a1 → a2 → b1 → is an Auslander-Reiten triangle.The distinguished triangle on the second row is given by Corollary 3.4.32.

Since a+ is indecomposable, the morphism Σ−1c+ → a2 is non-zero. Simi-larly, since a1 is indecomposable, the morphism a2 → b1 is non-zero, hencethe morphism Σ−1c+ → b1 is non-zero by Lemma 3.4.6.

Finally, the mapping cone of the morphism Σ−1c+ → b1 is b−, by a mirrorversion of Lemma 3.4.28.

Lemma 3.4.34. The mapping cone of the morphism m+ → b+ is d2.

Proof. The Hom space (m+, b+) is one-dimensional by Corollary 3.4.11. ByCorollary 3.4.32, the mapping cone of m2 → m+ is b+. Hence the mappingcone of m+ → b+ is Σm2 = d2.

Lemma 3.4.35. The mapping cone of the morphism a3 → a+ ⊕ b+ is d2.


115

m+

0 //

b+

b+

θ1

//

(i)

0

a3

// a+ ⊕ b+

// ∗ γ1 //

θ2

(ii)

e3

a3// a+

// d−γ2 // e3.

The distinguished triangle on the last row is given by Lemma 3.4.31 and thedistinguished triangle in the third column is given by Lemma 3.4.34.

Finally, rest reassured that the morphism m+ → b+ considered is non-zero.Assume it is. Then the object ∗ would have to be isomorphic to b+ ⊕ d−,and θ1 would have to be the non-zero canonical inclusion and θ2 would haveto be the non-zero canonical surjection. Since square (i) is commutative, γ1

would have to be zero although e3 is in R(b+). Similarly, since square (ii) iscommutative, γ2 would have to be zero although e3 is in R(d−). However,this is a contradiction since a+ is indecomposable. Therefore the morphismm+ → b+ considered is non-zero indeed.

Remark 3.4.36. In Lemma 3.4.35, it is not possible to use [16, Lemma 6.5]to show that ∗ is indecomposable, as in Lemma 3.4.28.

Corollary 3.4.37. The mapping cone of the morphism a2 → a+⊕ b+ is c3.


Σ−1c+

0 //

b+

b+

a2

// a+ ⊕ b+

// ∗ //

a2

// a+// c+

// .

The distinguished triangle on the last row is given by Lemma 3.4.32. There isan Auslander-Reiten triangle b+ → c3 → c+ →, and the result is immediatesince (a2, a+ ⊕ b+) is one-dimensional.

116

Finally, the morphism Σ−1c+ → b+ considered is non-zero indeed. This isvery similar to Lemma 3.4.35 where the morphism m+ → b+ is non-zero.

Lemma 3.4.38. The mapping cone of the morphism a− → b+ is e1.

Proof. By a mirror version of Corollary 3.4.33, the mapping cone of themorphism a1 → a− is b+. Hence the mapping cone of the morphism a− → b+is Σa1 = e1.

Using the same notation as in [23, Theorem 2.3], let σ be the translationfunctor of the triangulated category TX . The construction of the functor σis described in [23, Setup 1.1].

Lemma 3.4.39. (c.f. Lemma 3.2.13) Let a = (u, v) be an indecomposableobject of TX . Let p and q be integers and let λ = v − u. Then

σ(u, v) =

(p+ 4, q + 1) if (u, v) = (p, q)−,(u+ λ, v + (5− λ)) = (v, u+ 5) if 1 < λ < 4,(u+ 1, v + 4)− if λ = 1.

Proof. The first line is true by Lemma 3.4.15 and Lemma 3.4.38. The sec-ond line is true by Lemma 3.4.13, Lemma 3.4.14, Lemma 3.4.35 and Corol-lary 3.4.37. The last line is true by Lemma 3.4.16 and Corollary 3.4.33.

The action of σ is shown in the following diagram.

Diagram 3.4.40.

•

•

•

•

•

•

•

•a−

•b−

•

•

•

. . . •

??

GG

•a3

??

GG

•

??

GG

•c3

??

GG

•

??

GG

•

??

GG

•

. . .

•

??

•a2

??

•

??

•

??

•d2

??

•

??

•

??

•

•a1

??

•

??

•

??

•

??

•e1??

•

??

•

??

The action of σ is given by σ(a1) = b−, σ(a2) = c3, σ(a3) = d2 and σ(a−) =e1.

Corollary 3.4.41. (c.f. Corollary 3.2.15) Let a = (p, q)− be an indecom-posable object of TX , N ≥ 0 an integer. Then

σN (a) =

(p+ 5N

2 , q + 5N2 )− for N even,

(p+ 5(N+1)2 − 1, q + 5(N−1)

2 + 1) for N odd.

117


Corollary 3.4.42. (c.f. Corollary 3.2.15) Let a = (u, v) be an indecom-posable object of TX . Let λ = v − u, N ≥ 0 an integer. If 1 < λ < 4,then

σN (a) =

(u+ 5N

2 , v + 5N2 ) for N even,

(v + 5(N−1)2 , u+ 5(N+1)

2 ) for N odd.


Corollary 3.4.43. (c.f. Corollary 3.2.15) Let a = (u, v) be an indecompos-able object of TX . Let N ≥ 0 be an integer. If v − u = 1, then

σN (a) =

(u+ 5N

2 , v + 5N2 ) for N even,

(u+ 5(N−1)2 + 1, v + 5(N+1)

2 − 1)− for N odd.


For example, let a = (u, v) be an indecomposable object. Then σ2(u, v) =(u+ 5, v + 5).


Lemma 3.4.44. (c.f. Lemma 3.2.16) Let a = (u, v) be an indecomposableobject of TX . Let p and q be integers and let λ = v − u. Then

σ−1(u, v) =

(p− 1, q − 4) if (u, v) = (p, q)−,(v − 5, u) if 1 < λ < 4,(u− 4, v − 1)− if λ = 1.


Corollary 3.4.45. (c.f. Corollary 3.2.18) Let a = (p, q)− be an indecom-posable object of TX , N ≥ 0 an integer. Then

σ−N (a) =

(p− 5N

2 , q −5N2 )− for N even,

(p− 5(N−1)2 − 1, q − 5(N+1)

2 + 1) for N odd.


118

Corollary 3.4.46. (c.f. Corollary 3.2.18) Let a = (u, v) be an indecom-posable object of TX . Let λ = v − u, N ≥ 0 an integer. If 1 < λ < 4,then

σ−N (a) =

(u− 5N

2 , v −5N2 ) for N even,

(v − 5(N+1)2 , u− 5(N−1)

2 ) for N odd.


Corollary 3.4.47. (c.f. Corollary 3.2.18) Let a = (u, v) be an indecompos-able object of TX . Let N ≥ 0 be an integer. If v − u = 1, then

σ−N (a) =

(u− 5N

2 , v −5N2 ) for N even,

(u− 5(N+1)2 + 1, v − 5(N−1)

2 − 1)− for N odd.


Lemma 3.4.48. (c.f. Lemma 3.2.19) Let N ≥ 1 be an integer. Considera = a1 ⊕ a2 ⊕ a3 ⊕ a− in TX . Then HomTX (a, σNa) = 0.

Proof. Similar to Lemma 3.2.19, with the help of Lemma 3.4.18, Lemma 3.4.19,Lemma 3.4.20, Lemma 3.4.21, Corollary 3.4.41, Corollary 3.4.42 and Corol-lary 3.4.43.

Lemma 3.4.49. (c.f. Lemma 3.2.20) Let N ≥ 1 be an integer. Considera = a1 ⊕ a2 ⊕ a3 ⊕ a− in TX . Then HomTX (a, σ−Na) = 0.

Proof. Similar.

3.4.4 Theorem


Theorem 3.4.50. Consider the categories T , X and TX as usual, and theobject a = a1 ⊕ a2 ⊕ a3 ⊕ a− in TX as in Lemma 3.4.22. Then there is

an equivalence of triangulated categories f : TX'→Db(mod (kA4)) where

f(a) = (kA4).

Proof. Let U and X in Theorem 3.1.1 be TX and a respectively. By Corol-lary 3.4.24, A ∼= (kA4). Let us take Σ in Theorem 3.1.1 to be the transla-tion functor σ of the triangulated category TX . Then the result follows fromLemma 3.4.22, Lemma 3.4.48 and Lemma 3.4.49.

119

Chapter 4

A characterization of torsiontheories in the clustercategory of Dynkin type A∞

This chapter is also written in the form of a paper which is submitted forpublication in Homology, Homotopy and Applications ([37]).

4.1 Introduction

The cluster category D of Dynkin type A∞ was introduced in [19]. One ofits several definitions, which is completely analogous to the definition of thecluster category of type An, motivates us to say that D is a cluster categoryof type A∞. Namely, it is the orbit category Df (mod Γ)/SΣ−2. Here Γ isa quiver of type A∞ with zigzag orientation and S and Σ are the Serre andtranslation functors of the finite derived category Df (mod Γ).

There are also several other ways to realize the category D. In brief, it isthe algebraic triangulated category generated by a 2-spherical object. It isalso the compact derived category Dc(A) of the differential graded cochainalgebra A = C∗(S2; k) where S2 is the 2-sphere and k is a field. Finally, Dis the finite derived category Df (k[T ]) where k[T ] is viewed as a DG algebrawith T placed in homological degree 1 and zero differential. It is ubiquitousand the reader can refer to [19, Section 0] for more details.

In [19], the cluster tilting subcategories of D were shown to be in bijectionwith certain maximal sets of non-crossing arcs connecting non-neighbouringintegers. One can think of these maximal sets as “triangulations of the∞-gon”.

120

A torsion theory in D is a pair (X ,Y) of subcategories such that there areno non-zero morphisms from any object in X to any object in Y, and thatfor each d in D, there is a distinguished triangle x→ d→ y → in D, with xin X and y in Y (Definition 1.2.5). If T is a cluster tilting subcategory, then(T , ΣT ) is a torsion theory, but t-structures and co-t-structures are otherexamples of torsion theories.

In this chapter, the results of [19] are generalized by giving a bijection be-tween torsion theories in D and certain configurations of arcs connectingnon-neighbouring integers. A few examples, characterizing all t-structuresand co-t-structures in D, are given.

4.2 Coordinate system

Let us recollect some material from [19]. The category D has finite-di-mensional Hom spaces over a field k and split idempotents, so it is Krull-Schmidt ([19, Remark 1.2]). In this chapter, a subcategory of D is assumedto be a full subcategory closed under direct sums and direct summands. Forsubcategories X and Y of D, the set of morphisms from any x in X to anyy in Y is denoted by (X ,Y). In particular, for objects x and y of D, the setof morphisms from x to y is denoted by (x, y).

By [19, Remark 1.4], the Auslander-Reiten quiver of D is ZA∞, and thefollowing standard coordinate system is laid down on the Auslander-Reitenquiver.

...

...

...

...

· · ·

??

(−4, 1)

??

(−3, 2)

??

(−2, 3)

??

· · ·

(−4, 0)

??

(−3, 1)

??

(−2, 2)

??

(−1, 3)

??

· · ·

??

(−3, 0)

??

(−2, 1)

??

(−1, 2)

??

· · ·

(−3,−1)

??

(−2, 0)

??

(−1, 1)

??

(0, 2)

??

Let Σ be the translation functor of D. Since D is 2-Calabi-Yau, its Serrefunctor is S = Σ2 and the Auslander-Reiten translation is τ = SΣ−1 = Σ. Interms of coordinates, the action of Σ = τ is given by Σ(m,n) = (m−1, n−1),see [19, Remark 1.4].

The definitions to follow interpret coordinate pairs in an alternative way, asarcs connecting non-neighbouring integers.

Definition 4.2.1. ([19, Definition 3.1]) An arc is a pair (m,n) of integerswith n−m ≥ 2. The arc (m,n) is said to end in each of the integers m and n.

121

Two arcs (m1, n1) and (m2, n2) are said to cross if eitherm1 < m2 < n1 < n2

or m2 < m1 < n2 < n1. The action of Σ makes sense on arcs as well.

In the diagrams to follow arcs are drawn on number lines which are numberedthus.

−2 −1 0 1 2 3

For example, in the above diagram, the arc (−1, 2) is drawn as a curvebetween the integers −1 and 2. Crossing of arcs has been defined to matchthis geometrical picture in the natural way.

This, together with the following definition, relies heavily on the coordinatesystem.

Definition 4.2.2. ([19, Definition 3.2]) Let A be a set of arcs. If, for eachinteger n, there are only finitely many arcs in A which end in n, then A issaid to be locally finite. A left (resp. right) fountain of A is an integer nfor which there are infinitely many arcs of the form (m,n) (resp. (n,m)) inA. A fountain of A is an integer n which is both a left and a right fountainof A. A left (resp. right) fountain n of A is said to be full if for each mthe arc of the form (m,n) (resp. (n,m)) is in A. Finally, A is said to benon-crossing if A does not contain any pairs of crossing arcs.

Coordinate pairs (m,n), n−m ≥ 2, are identified with either indecomposableobjects of D or with arcs. Thereupon a set A of arcs induces a collection ofindecomposable objects of D, and add of them gives a subcategory A of D,thus there is a bijection between sets of arcs and subcategories of D.

It is our intention to display (in diagrams) at times both versions of interpre-tations, coordinate pairs as indecomposable objects of D on the quiver andas arcs on number lines. The reader should be able to feel so comfortablewith the simultaneous visualizations in the mind’s eye, so much so that anyone interpretation is at once able to find an echo and a corroboration in theother.

The following pairs of definitions describe some constraints on a set of arcsA.

Definition 4.2.3. A set of arcs A is said to satisfy condition (?) if, for eachpair of crossing arcs (a, b) and (c, d) in A, those of the pairs (a, c), (c, b),(b, d) and (a, d) which are arcs belong to A (for instance, (a, c) is only anarc if c− a ≥ 2).

122

a c b d

The following example visualizes condition (?) on the quiver.

Example 4.2.4. Suppose the set of arcs A satisfies condition (?). Then giventhe arcs (−4, 0) and (−2, 2) in A, the arcs (−4,−2), (−2, 0), (0, 2) and (−4, 2)are in A.

...

...

...

...

...

· · ·

??

(−6, 1)

??

(−5, 2)

??

(−4, 3)

??

(−3, 4)

??

· · ·

(−6, 0)

??

(−5, 1)

??

(−4, 2)

??

(−3, 3)

??

(−2, 4)

??

· · ·

??

(−5, 0)

??

(−4, 1)

??

(−3, 2)

??

(−2, 3)

??

· · ·

(−5,−1)

??

(−4, 0)

??

(−3, 1)

??

(−2, 2)

??

(−1, 3)

??

· · ·

??

(−4,−1)

??

(−3, 0)

??

(−2, 1)

??

(−1, 2)

??

· · ·

(−4,−2)

??

(−3,−1)

??

(−2, 0)

??

(−1, 1)

??

(0, 2)

??

Definition 4.2.5. A set of arcs A is said to satisfy condition (? ?) if it hasthe following property: if a is a left fountain but not a right fountain of A,b is a right fountain but not a left fountain of A and b− a ≥ 2, then the arc(a, b) is in A.

a b

The following example visualizes condition (? ?) on the quiver.

Example 4.2.6. Suppose the set of arcs A satisfies condition (? ?). Thengiven the left fountain (but not right fountain) −1 and the right fountain(but not left fountain) 2 of A, the arc (−1, 2) is in A. The diagram belowshows only some of the arcs from the fountains.

123

...

...

...

...

...

...

...

...

· · ·

??

(−6, 1)

??

(−5, 2)

??

(−4, 3)

??

(−3, 4)

??

(−2, 5)

??

(−1, 6)

??

(0, 7)

??

· · ·

(−6, 0)

??

(−5, 1)

??

(−4, 2)

??

(−3, 3)

??

(−2, 4)

??

(−1, 5)

??

(0, 6)

??

(1, 7)

??

· · ·

??

(−5, 0)

??

(−4, 1)

??

(−3, 2)

??

(−2, 3)

??

(−1, 4)

??

(0, 5)

??

(1, 6)

??

· · ·

(−5,−1)

??

(−4, 0)

??

(−3, 1)

??

(−2, 2)

??

(−1, 3)

??

(0, 4)

??

(1, 5)

??

(2, 6)

??

· · ·

??

(−4,−1)

??

(−3, 0)

??

(−2, 1)

??

(−1, 2)

??

(0, 3)

??

(1, 4)

??

(2, 5)

??

· · ·

(−4,−2)

??

(−3,−1)

??

(−2, 0)

??

(−1, 1)

??

(0, 2)

??

(1, 3)

??

(2, 4)

??

(3, 5)

??

Definition 4.2.7. Given a subcategory U of D, let U⊥ = d ∈ D|(u, d) = 0for all u in U and ⊥U = d ∈ D|(d, u) = 0 for all u in U.

Definition 4.2.8. Given a set of arcs U, let ort(U) be the set of arcs ort(U)= d | d does not cross any arcs in U.

Lemma 4.2.9. Let U be a set of arcs. Then ort ort ort(U) = ort(U).

Proof. That ort ort(U) ⊇ U is immediate, so that ort ort ort(U) ⊆ ort(U),since ort reverses inclusions. Similarly, ort ort ort(U) ⊇ ort(U). Thereforeort ort ort(U) = ort(U).

Lemma 4.2.9 gives the minimal level when the operation ort becomes peri-odic. Later on, Lemma 4.4.20 gives the equivalent condition for the equalityort ortA = A.

Remark 4.2.10. In addition, the operation ort2 satisfies the following.

(i) ort ort(U) ⊇ U,

(ii) ort4(U) = ort2(U), which is immediate from Lemma 4.2.9.

Lemma 4.2.11. Let U be a set of arcs. Then U = ort(C) for some set ofarcs C if and only if U = ort ort(U).

Proof. (if ) This is immediate. (only if ) Since U = ort(C), therefore ort ort(U) =ort ort ort(C) = ort(C) = U by Lemma 4.2.9.

Lemma 4.2.12. Let U, U′ be sets of arcs. Suppose ort ort(U) = U, ort ort(U′) =U′. Then

(i) If ort(U) ⊇ ort(U′), then U ⊆ U′.

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(ii) If ort(U) = ort(U′), then U = U′.

Proof. (i) If ort(U) ⊇ ort(U′), then ort ort(U) ⊆ ort ort(U′) which givesU ⊆ U′.

(ii) This is immediate by (i).

Lemma 4.2.13. Let U be a set of arcs. Then

(i) U is non-crossing if and only if U ⊆ ort(U).

(ii) U is maximal non-crossing if and only if U = ort(U).

Proof. This is immediate.

Let x = (i, j) be an indecomposable object of D, accompanied by the regionsH−(x) = (m,n) | m ≤ i − 1, i + 1 ≤ n ≤ j − 1 and H+(x) = (m,n) |i + 1 ≤ m ≤ j − 1, j + 1 ≤ n in the Auslander-Reiten quiver of D ([19,Definition 2.1]). They are sketched as follows.

H−(x) H+(x)

Σx x Σ−1x

We write H(x) = H−(x) ∪H+(x).

Remark 4.2.14. Let (a, b) and (c, d) be arcs. Then the two arcs cross if andonly if (c, d) is in H(a, b).

The following lemma describes morphisms in the category D.

Lemma 4.2.15. ([19, Corollary 2.3]) Let x and y be indecomposable objectsof D. Then the following are equivalent.

(i) (x, y) 6= 0,

(ii) (x, y) = k,

(iii) y ∈ H(Σx),

(iv) x ∈ H(Σ−1y).

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Remark 4.2.16. By virtue of the coordinate system, the way the regionsH−(x) and H+(x) are defined, and the above lemma, crossing of arcs is en-dowed with a meaningful interpretation. This is to say, given indecompos-able objects x and y ofD, then (x, y) 6= 0 if and only if the arcs correspondingto x and Σ−1y cross, see [19, Lemma 3.6].

Now let X and Y be two subcategories of D such that X = ⊥Y and Y = X⊥,accompanied by the sets of arcs X and Y respectively. By Remark 4.2.16,X = d | Σd does not cross any arcs in Y and Y = d | Σ−1d does not crossany arcs in X. Together they determine each other.

Let Σ−1Y = W. The above can be rewritten as X = d | d does not cross anyarcs in W = ort(W) and W = d | d does not cross any arcs in X = ort(X).

The following lemma bridges indecomposable objects of D on the quiver andarcs on number lines.

Lemma 4.2.17. (double orthogonal property) Let X and X be defined asabove. Then X = ⊥(X⊥) is equivalent to X = ort ort(X).

Proof. This is immediate by the above description.

4.3 Precovering (preenveloping) subcategories

In this section, precovering and preenveloping subcategories are character-ized in terms of their corresponding sets of arcs.

Theorem 4.3.1. Let A be a subcategory of D and let A be the correspondingset of arcs. Then A is precovering if and only if each right fountain of A isin fact a fountain.

Proof. Suppose A is precovering. A right fountain of A is an integer n forwhich there are infinitely many arcs of the form (n, p) in A. The correspond-ing collection P of indecomposable objects of A lies on a diagonal half liner in the Auslander-Reiten quiver of D. The following sketch shows r alongwith some of the indecomposable objects ai in P (indicated by the blackdots) and, in dotted lines, their respective regions H(Σai).

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s′ r

•s

••

To show that n is also a left fountain, that is, there are infinitely many arcsin A of the form (m,n), is the same as showing that there are infinitelymany indecomposable objects of A which are on the half line s. Consideran object x on the half line s and its region H(Σ−1x) indicated by dashedlines in the following diagram.

l3 t′ s′ r

l2l1

•s

•x •

Let β : b → x be an A-precover, where b and β are written b1 ⊕ · · · ⊕ bqand

(β1, . . . , βq

)respectively. The morphism β : b → x can be assumed to

be non-zero on each direct summand bj of b, so that all the bj belong toH(Σ−1x).

It is apparent that x is in H−(Σai) for all the ai in P . Pick an indecomposablea in P and let α : a→ x be a non-zero morphism. Then α factors through β

as shown in the following diagram, where the morphism γ is written

γ1...γq

.

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a

α

''

γ

b1 ⊕ · · · ⊕ bq

β// x

Since α is non-zero and α = β1γ1 + · · ·+ βqγq, there is a term βkγk which isnon-zero. Hence γk : a→ bk is non-zero so bk is in H(Σa).

Therefore bk can only be on the half line s above x, or in the infinite regioninside H+(Σ−1x) bounded by t′, s′ and li for some i. But it cannot be thelatter because then βk and γk would both be backward morphisms whenceβkγk would be zero (Lemma 3.3.9). Therefore bk can only be on the half lines above x. Repeating the argument, consider an object x1 on the half line sabove bk. Then another object c in A on the half line s above x1 is revealed.Since it can be continued in this way indefinitely, there are infinitely manyindecomposable objects of A which are on the half line s.

Now suppose each right fountain of A is in fact a fountain.

In the following diagram, consider an object x and its region H(Σ−1x) inthe Auslander-Reiten quiver of D, indicated by wavy lines. As indicated,half lines (which start from the bottom line) of the form ri or si, wherei ∈ N is a variable, are also introduced. Suppose the region H−(Σ−1x)(resp. H+(Σ−1x)) has boundary half lines s0 and sm (resp. r0 and rm).Then each line si (resp. ri), 0 ≤ i ≤ m, has to pass through the regionH−(Σ−1x) (resp. H+(Σ−1x)) and is parallel to the boundary lines of theregion H−(Σ−1x) (resp. H+(Σ−1x)).

sm r0

ri

si x

s0 rm

t

Let S be the intersection of H−(Σ−1x) and the objects of A. On eachline si, 0 ≤ i ≤ m, consider the first object ai in A which lies above theline segment t. Denote by as the direct sum of all the ai and consider the

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canonical morphism as → x. By Lemma 3.3.6, each morphism a → x witha in S factors as a→ as → x.

For example, in the following diagram, let m = 5 and the circles and bulletsindicate the first few objects of A on each line si. The object as describedabove is the direct sum of the objects indicated by circles .

s5

s4

s3

s2 •s1 x

s0 • • t

•

Let R be the intersection of H+(Σ−1x) and the objects of A. There is thedesire of an ar in A with a morphism ar → x such that each morphisma→ x with a in R factors as a→ ar → x.

Suppose R is finite. Then let ar be the direct sum of the objects in R(Lemma 2.5.1). Otherwise R is infinite. Since there are only finitely manyri, there is a line rj which contains infinitely many objects in A with a non-zero morphism to x. Let J ⊆ 0, . . . ,m consist of the j such that the linerj contains infinitely many objects in A with a non-zero morphism to x.

Now for each j in J , the line rj corresponds to a right fountain of A, and byassumption, each right fountain is also a fountain. Hence the correspondingline sj contains infinitely many objects in A with a non-zero morphism to x.Among the sj with j in J , take the line sq which is closest to the boundaryline sm, and then consider any object which is in both A and H−(Σ−1x) onthat line. By Lemma 3.3.4, this object plays the role of ar.

For example, in the following diagram, m = 5 and J = 1, 2. The line sqdescribed above is the line s2 here. The bullets on the ri indicate the firstfew objects of A in H+(Σ−1x). The bullets and the circle on the si indicatethe first few objects of A on those half lines. The object ar described aboveis indicated by the circle (one of the infinitely many choices).

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s5 r0

r1

• r2

s2

s1 x • •s0 • • • r5

• •

•

Finally, an A-precover of x can be obtained as ar ⊕ as → x.

Remark 4.3.2. (i) In the only if part of the proof of Theorem 4.3.1, theway the position of the direct summand bk is calibrated is very similarto the descriptions found in two previous situations, by simply lookingat which morphisms are zero and which are not. The first situation isin Lemma 3.2.7, where the regions LX (a) and RX (a) are determined.The other situation is in Lemma 3.4.28, where the mapping cone ofthe morphism µ : a+ → d1 is determined.

(ii) In the if part of the proof of Theorem 4.3.1, the occurrence of the back-ward morphisms begets the simultaneous occurrence of a left fountain(where it is a right fountain already) in pursuance of symmetry. Eventhough the introduction of the left fountain induces more objects inA, it also provides another distribution of objects in A with differentsuggestiveness, thus permitting A to be precovering.

The following is the dual of Theorem 4.3.1.

Theorem 4.3.3. Let A be a subcategory of D and let A be the correspondingset of arcs. Then A is preenveloping if and only if each left fountain of A isin fact a fountain.

Proof. Similar.

4.4 Torsion theories

In this section, let A be a set of arcs. A sequence of results regarding Ais given, and then also a checkable condition equivalent to X = ort ort(X),see Lemma 4.4.20. Finally, the main theorem of this chapter is given inTheorem 4.4.22.

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Definition 4.4.1. Let U be a subcategory of D. Then U is said to be weakcluster tilting if it satisfies U = (Σ−1U)⊥ and U = ⊥(ΣU), and is said to becluster tilting if it is weak cluster tilting, precovering and preenveloping.

In [19], the cluster tilting subcategories of D were shown to be in bijectionwith certain maximal sets of non-crossing arcs connecting non-neighbouringintegers. This is rephrased as follows.

Theorem 4.4.2. ([19, Theorem 4.4]) Let U be a weak cluster tilting sub-category of D with U as the corresponding maximal set of non-crossing arcs.Then U is precovering and preenveloping (that is, U is a cluster tilting sub-category of D) if and only if U is (i) locally finite, or (ii) has a fountain.

In this section, the above theorem is generalized in terms of a bijectionbetween torsion theories in D and certain configurations of arcs connectingnon-neighbouring integers.

The following little lemma is a simple play of concepts.

Lemma 4.4.3. Let (a, b) be an arc (not necessarily in A). Then the follow-ing are equivalent.

(i) ort ortA = A,

(ii) If (a, b) is an arc such that each arc crossing (a, b) also crosses an arcin A, then (a, b) is in A.

Proof. (i)⇒ (ii): Suppose ort ortA = A and let the arc (a, b) be as described.Then the arc (a, b) is in ort ortA, otherwise there would have to be an arc(c, d) in ortA which crossed (a, b). Subsequently, (c, d) would be an arc whichcrossed (a, b) but would not cross any arcs in A, which is a contradiction.Therefore the arc (a, b) is in ort ortA = A.

(ii) ⇒ (i): Suppose otherwise that ort ortA ⊃ A. Let (a, b) be an arc inort ortA but not in A. Then there has to be an arc (c, d) which crosses(a, b) but does not cross any arcs in A. Therefore (c, d) is in ortA. However,(c, d) is not in ort ort ort(U), which is a contradiction since ort ort ort(U)= ort(U) by Lemma 4.2.9. Therefore ort ortA ⊆ A which, together withort ort(U) ⊇ U, gives ort ortA = A.

Now let us enter the spirit of this section.

Lemma 4.4.4. Suppose ort ortA = A. Then A satisfies conditions (?) and(? ?) of Definitions 4.2.3 and 4.2.5.

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Proof. To see that A satisfies condition (?), consider the diagram in Defini-tion 4.2.3. The arc (a, c) is in ort ortA, otherwise there would have to bean arc (m,n) in ortA which crossed (a, c), but the arc (m,n) cannot be inortA since it crosses either the arc (a, b) or the arc (c, d) in A. Thereforethe arc (a, c) is in ort ortA = A. The rest is similar.

To see that A satisfies condition (? ?), consider the diagram in Defini-tion 4.2.5. Suppose the arc (a, b) is not in ort ortA. Then there is an arc(m,n) in ortA which crosses (a, b), that is, m < a < n < b or a < m < b < n.For the first case the arc (m,n) cannot however be in ortA, since there isalways an arc (q, a) with q < m in A which crosses (m,n), and similarly forthe second case. Therefore the arc (a, b) is in ort ortA = A.

Lemma 4.4.5. Suppose A satisfies condition (?) and suppose there are onlyfinitely many (but not zero) arcs in A that end in a. Suppose there are botharcs going to the left and arcs going to the right from a. If (p, a) is thelongest arc in A going to the left from a and (a, q) is the longest arc in Agoing to the right from a, then (p, q) is an arc in ortA.

p a q

Proof. There are no arcs (m,n) in A withm < p, p < n < a, otherwise (m, a)would be in A by condition (?), contradicting that (p, a) is the longest arc inA going to the left from a. There are also no arcs (m,n) in A with p < m < a,q < n, otherwise (a, n) would be in A by condition (?), contradicting that(a, q) is the longest arc in A going to the right from a.

Similarly, there are no arcs (m,n) in A with a < m < q, q < n, and thereare also no arcs (m,n) in A with m < p, a < n < q. By construction thereare no arcs (m, a) in A with m < p, and there are also no arcs (a, n) in Awith q < n.

Combining all these shows that there are no arcs in A crossing (p, q) so (p, q)has to be in ortA.

Lemma 4.4.6. Suppose A satisfies condition (?) and suppose there are onlyfinitely many (but not zero) arcs in A that end in a. Suppose there are onlyarcs going to the left from a. If (p, a) is the longest arc in A going to theleft from a, then (p, a+ 1) is in ortA.

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p a a+ 1

Proof. There are no arcs (m,n) in A with m < p, p < n < a, otherwise(m, a) would be in A by condition (?), contradicting that (p, a) is the longestarc in A going to the left from a. There are also no arcs (m,n) in A withp < m < a, n > a + 1, otherwise (a, n) would be in A by condition (?),contradicting that there are no arcs in A going to the right from a. Byconstruction there are no arcs (m, a) in A with m < p and it is a conditionthat there are no arcs (a, n) in A. Therefore (p, a+1) has to be in ortA.

Lemma 4.4.7. Suppose A satisfies condition (?) and suppose there are onlyfinitely many (but not zero) arcs in A that end in a. Suppose there are onlyarcs going to the right from a. If (a, p) is the longest arc in A going to theright from a, then (a− 1, p) is in ortA.

a− 1 a p


Remark 4.4.8. Suppose (a, b) is an arc in ort2(A). Then there has to besome arc in A which ends in a (resp. b). Otherwise the arc (a − 1, a + 1)(resp. (b−1, b+1)) is in ortA, which is a contradiction since it crosses (a, b)in ort2(A).

Corollary 4.4.9. Suppose A satisfies condition (?). Suppose that (a, b) isan arc in ort2(A) and that there are only finitely many arcs in A that endin a. Then it is not possible that all arcs in A that end in a are of the form(m, a).

p a b

Proof. Suppose all arcs in A that end in a were of the form (m, a) and let(p, a) be the longest one. Then by Lemma 4.4.6 the arc (p, a + 1) is in ortA, but this is a contradiction since it crosses (a, b) which is in ort2(A).

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Corollary 4.4.10. Suppose A satisfies condition (?). Let (a, b) be an arcin ort2(A) and suppose there are only finitely many arcs in A that end in a.Then there is an arc (a,m) in A with b ≤ m.

Proof. By Corollary 4.4.9, it is not possible that all arcs in A that end in aare of the form (m, a). Therefore only the following two cases remain.

(i) Suppose there are arcs in A going to the left and going to the rightfrom a. Let (a, q) be the longest arc in A going to the right from a and(p, a) be the longest arc in A going to the left from a. By Lemma 4.4.5,(p, q) is in ortA, so q < b is not possible since then (p, q) would cross(a, b) in ort2(A).

p a q b

Therefore b ≤ q.

(ii) Suppose there are only arcs in A going to the right from a. Let (a, p)be the longest arc in A going to the right from a. By Lemma 4.4.7,(a−1, p) is in ortA, so p < b is not possible since then (a−1, p) wouldcross (a, b) in ort2(A).

a− 1 a p b

Therefore b ≤ p.

Corollary 4.4.11. Suppose A satisfies condition (?). Suppose that (a, b) isan arc in ort2(A) and that there are only finitely many arcs in A that endin b. Then it is not possible that all arcs in A that end in b are of the form(b,m).

Proof. Similar to Corollary 4.4.9.

Corollary 4.4.12. Suppose A satisfies condition (?). Let (a, b) be an arcin ort2(A) and suppose there are only finitely many arcs in A that end in b.Then there is an arc (m, b) in A with m ≤ a.

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Proof. Similar to Corollary 4.4.10.

Lemma 4.4.13. Suppose A satisfies condition (?). Let (a, b) be an arc inort2(A) and suppose that each of a and b is only an end point of finitelymany arcs in A. Then (a, b) is in A.

Proof. By Corollary 4.4.10, there is an arc (a, q) in A with b ≤ q. On theother hand by Corollary 4.4.12, there is an arc (p, b) in A with p ≤ a. Ifq = b or p = a already, then this is what is to be shown. Assume otherwisethat q > b and p < a. But then (a, b) is in A by condition (?).

p a b q

Lemma 4.4.14. Suppose A satisfies condition (?). Let both a and b be right(resp. left) fountains of A with b− a ≥ 2. Then (a, b) is in A.

Proof. Suppose both a and b are right fountains of A. Choose an arc (a, p)in A with b < p and then choose an arc (b, q) in A with p < q. Then (a, b)is in A by condition (?).

a b p q

The other case is similar.

Lemma 4.4.15. Suppose A satisfies condition (?). Let a be a right fountainof A, b be a left fountain of A with b− a ≥ 2. Then (a, b) is in A.

Proof. Choose an arc (a, q) in A with b < q and then choose an arc (p, b) inA with p < a. Then (a, b) is in A by condition (?).

p a b q

Lemma 4.4.16. Suppose A satisfies condition (?). Let (a, b) be an arc inort2(A). Suppose b is a left fountain of A and suppose there are only finitelymany arcs in A that end in a. Then (a, b) is in A.

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Proof. By Corollary 4.4.10, there is an arc (a,m) in A with b ≤ m. If m = balready, then this is what is to be shown. Otherwise choose an arc (p, b) inA with p < a. Then (a, b) is in A by condition (?).

p a b m

Lemma 4.4.17. Suppose A satisfies condition (?). Let (a, b) be an arc inort2(A). Suppose b is a right fountain of A and suppose there are only finitelymany arcs in A that end in a. Then (a, b) is in A.

Proof. By Corollary 4.4.10, there is an arc (a,m) in A with b ≤ m. If m = balready, then this is what is to be shown. Otherwise choose an arc (b, q) inA with m < q. Then (a, b) is in A by condition (?).

a b m q

Lemma 4.4.18. Suppose A satisfies condition (?). Let (a, b) be an arcin ort2(A). Suppose a is a right fountain of A and suppose there are onlyfinitely many arcs in A that end in b. Then (a, b) is in A.


Lemma 4.4.19. Suppose A satisfies condition (?). Let (a, b) be an arc inort2(A). Suppose a is a left fountain of A and suppose there are only finitelymany arcs in A that end in b. Then (a, b) is in A.


Finally, we deliver the following lemma which is a recollection of the abovelemmas.

Lemma 4.4.20. (c.f. Lemma 4.2.17) ort ortA = A if and only if A satisfiesconditions (?) and (? ?).

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Proof. (only if ) This is Lemma 4.4.4. (if ) It is immediate that A ⊆ ort ortA.Let (a, b) be an arc in ort2(A). Suppose that each of a and b is only an endpoint of finitely many arcs in A. Then (a, b) is in A by Lemma 4.4.13.Otherwise suppose that both a and b are end points of infinitely many arcsin A. Then (a, b) is in A by Lemma 4.4.14, Lemma 4.4.15 and condition (? ?).Finally, suppose that precisely one of a and b is an end point of finitely manyarcs in A. Then (a, b) is in A by Lemma 4.4.16, Lemma 4.4.17, Lemma 4.4.18and Lemma 4.4.19.

Unlike Lemma 4.4.3, Lemma 4.4.20 expresses explicitly ort ortA = A interms of the configuration of A. It can also be seen directly that the conditionin Lemma 4.4.3 does imply conditions (?) and (? ?).

Example 4.4.21. (i) Let U be a set of arcs. By Lemma 4.2.9, ort ort ort(U)= ort(U). Therefore ort(U) satisfies conditions (?) and (? ?) by Lemma 4.4.20.The same is true for ortn(U), n ≥ 1.

(ii) Suppose A is a set of non-crossing arcs where ort ortA = A. Thenit is not necessary that A is a maximal set of non-crossing arcs. Forexample, let A consist only of a left fountain a, a right fountain b andthe arc (a, b). The following diagram shows only some of the arcs fromthe configuration described.

a b

By Lemma 4.4.20, ort ortA = A. However, it is certainly not a maxi-mal set of non-crossing arcs. For example, it is possible to add an arcas in the following diagram.

a b

Now we are ready to deliver the main theorem of this chapter.

Theorem 4.4.22. Let X be a subcategory of D and let X be the correspond-ing set of arcs. Then the following conditions are equivalent.

(i) X satisfies conditions (?) and (? ?), and each right fountain of X is infact a fountain,

(ii) The subcategory X is precovering and is closed under extensions,

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(iii) (X ,Y) is a torsion theory for some subcategory Y of D. In particular,Y = X⊥.

Proof. (i) ⇒ (ii): X satisfying conditions (?) and (? ?) implies X beingclosed under extensions by Lemma 4.2.17 and Lemma 4.4.20 (as well asLemma 1.2.19). Finally, X is precovering if and only if each right fountainof X is in fact a fountain by Theorem 4.3.1. (ii) ⇔ (iii): This is true by [21,Proposition 2.3]. (iii) ⇒ (i): Since (X ,Y) is a torsion theory, X = ⊥(X⊥)(Lemma 1.2.7) and X is precovering. Therefore X satisfies conditions (?)and (? ?) by Lemma 4.2.17 and Lemma 4.4.20 and each right fountain of Xis in fact a fountain by Theorem 4.3.1.

Example 4.4.23. (c.f. Example 4.4.21(ii)) In Theorem 4.4.22, if X is onlyclosed under extensions, then X does not necessarily satisfy condition (?).For example, by Lemma 3.3.18, (−4,−1)→ (−4, 1)→ (−2, 1)→ is a distin-guished triangle. Suppose the arcs (−4,−1), (−4, 1) and (−2, 1) are all inX. This however does not imply the arcs (−4,−2) and (−1, 1) in X, whichis required by condition (?).

−4 −3 −2 −1 0 1 2 3

Example 4.4.24. Let U be a cluster tilting subcategory with the correspond-ing set of arcs U. Then by definition (U , ΣU) is a torsion theory. Thereforeby Theorem 4.4.22, U satisfies conditions (?) and (? ?), and each right foun-tain of U is in fact a fountain. This is compatible with the description givenin Theorem 4.4.2, and is altogether a generalization of it.

Remark 4.4.25. This is a little digression. The way torsion theories general-ize cluster tilting subcategories reminds us of the way distinguished trianglesgeneralize Auslander-Reiten triangles. It is good mathematics to consider(different) special cases of the given notions, and then unveil the specialrelationships hidden not because they are particularly enigmatic but thatcoincidences are simply nature’s camouflage.

4.5 Examples

In this section, two special types of torsion theories in D are described, thoseof t-structures and co-t-structures. Given a subcategory U of D, exampleswhere the pair (U ,U⊥) might or might not be a torsion theory are given.In the context of a t-structure, an example is given to illustrate the torsiontheory triangles explicitly on the Auslander-Reiten quiver.

Let us first describe t-structures.

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Theorem 4.5.1. Let (U ,V) be a t-structure in D, that is, (U ,V) is a torsiontheory with ΣU ⊆ U . Suppose U is neither zero nor all of D, then there is ahalf line such that the indecomposable objects of U are precisely the objectson the half line and to the left of it, as shown in the following diagram.

U

Proof. Let U be the corresponding set of arcs for the subcategory U .

(Step 1 ) Consider a horizontal line y − x = k with k ≥ 3 in the Auslander-Reiten quiver of D. If there are objects from U on this line, then there isa rightmost such object. Namely, suppose not. Then there are objects ofU arbitrarily far to the right on y − x = k, so all objects on y − x = kare in U because ΣU ⊆ U . In the following diagram, let d1 = (u1, u2) andd2 = (u1 + 1, u2− 1) be objects on the lines y− x = k+ 1 and y− x = k− 1respectively. Then there is the Auslander-Reiten triangle d0 → d1 ⊕ d2 →d′0 →, where d0 = (u1, u2 − 1) and d′0 = (u1 + 1, u2). Since d0 and d′0both lie on the line y − x = k which is in U , it follows that d1 and d2 arein U , since U is closed under extensions and direct summands. Thereforethe two neighbouring lines y − x = k + 1 and y − x = k − 1 are in U .Repeating the argument for other (horizontal) lines, U has to contain allthe indecomposable objects of D, i.e. U has to be all of D. The case wherek = 2 is similar.

d1

d0

??

d′0

d2

??

(Step 2 ) Pick an object d = (m,n) in U and assume it is the rightmostobject of U on the horizontal line y−x = n−m. Here it will be shown thatthe region L = (x, y)|y ≤ n, y − x ≥ 2 is in U .

(i) Suppose n − m = 2. In the following diagram, let u1 = (m − 1, n).Then there is the Auslander-Reiten triangle Σd → u1 → d →. SinceΣU ⊆ U , therefore Σd is in U . Hence u1 is in U , since U is closed

139

under extensions. By applying a (similar) argument on u1 and so on,the half line y = n is in U , and so are all the half lines y = n′ withn′ ≤ n, since ΣU ⊆ U . Therefore the region L is in U .

y = n

u1

Σd d

(ii) Suppose n−m > 2. Since ΣU ⊆ U , therefore Σd = (m− 1, n− 1) is inU . In the following diagram, let u1 = (m− 1, n) and u2 = (m,n− 1).Then there is the Auslander-Reiten triangle Σd → u1 ⊕ u2 → d →.Therefore u1 and u2 are both in U , since U is closed under extensions.By applying a (similar) argument on u1 and u2 and so on (if possible),the two half lines t1 and t2 are in U . Eventually, the region labelledL0 (including the two half lines t1 and t2) is in U , since ΣU ⊆ U .

t1

u1

L0 Σd d

u2

t2 L1

Now it remains to show that the little triangular region, L1 = (x, y) |m+ 1 ≤ x ≤ n− 2,m+ 3 ≤ y ≤ n and y − x ≥ 2, is also in U .

Let r0 = (m − 1,m + 1). Since the arcs r0 and d = (m,n) cross, itfollows that q0 = (m + 1, n) is in U since U satisfies condition (?) byTheorem 4.4.22.

m− 1 m m+ 1 n

Similarly, with the help of r1 = (m − 1,m + 2), it follows that q1 =(m+ 2, n) is in U , and so on (if possible), until all the objects (m′, n),

140

with m+ 1 ≤ m′ ≤ n− 2, are in U . This argument is to be repeated,starting with u2 = (m,n − 1), u3 = (m,n − 2) and so on instead (ifpossible), until the region L1 is in U . Therefore L = L0 ∪ L1 is in U ,and this is what is needed to be shown.

u1

Σd d

u2 q0

u3 q1

r1

r0

(Step 3 ) Suppose there is an indecomposable object d′ = (u, v) of U whichlies outside the region L, i.e. n < v. Similar to Step 2, all the half linesy = v′ with v′ ≤ v will be in U . Therefore the indecomposable object Σ−1d =(m+ 1, n+ 1) will also be in U , contradicting that d is the indecomposableobject of U which is rightmost on the line y − x = n − m. Therefore theindecomposable objects of U are precisely the objects on the half line y = nand to the left of it, i.e. the region L.

Remark 4.5.2. Consider U and U in Theorem 4.5.1. Other possible choices ofcrossing of arcs within U would not give us any more new arcs not containedin U already (Example 4.2.4 and Remark 4.2.14).

Alternatively, one can conceive the set of arcs U as a sequence of left fullfountains going to the left. This means there is an integer p such that q is aleft full fountain of U for all q ≤ p and there are no left (full) fountains q > pof U (p is the rightmost left full fountain of U). Any arcs (a, b) induced bycrossing of arcs within U (those arcs that are to be included under condition(?)) are in U already (i.e. b has to be a left full fountain of U already).Therefore the set of arcs U in Theorem 4.5.1 does satisfy condition (?).

For example, let p = 2. The following diagram shows, amongst other arcsof U, the arcs (−4, 0) and (−2, 2) and they cross.

−4 −3 −2 −1 0 1 2

141

The arc (0, 2) has to be in U since 2 is a left full fountain, and similarly thearcs (−2, 0), (−4,−2) and (−4, 2) have to be in U, since 0, −2 and 2 are allleft full fountains.

Example 4.5.3. Let U be a subcategory of D. Lemma 4.2.15 is needed forthe following.

(i) In the following diagram, suppose the indecomposable objects of U areprecisely the objects on the dotted half line l1 : y = m (some m) andto the left of it. Then the indecomposable objects of U⊥ are preciselythe objects on the dotted half line l2 : x = m − 1 and to the right ofit, and the indecomposable objects of (U⊥)⊥ are precisely the objectson the dotted half line l3 : y = m− 2 and to the left of it.

l3 l1 l2

(U⊥)⊥ U U⊥

(ii) In the following diagram, suppose the indecomposable objects of U areprecisely the objects on the dotted half line l1 : y = m (some m) andto the left of it. Then the indecomposable objects of ⊥U are preciselythe objects on the dotted half line l2 : x = m + 1 and to the right ofit, and the indecomposable objects of ⊥(⊥U) are precisely the objectson the dotted half line l3 : y = m+ 2 and to the left of it.

l1 l3 l2

U ⊥(⊥U) ⊥U

Here it is true that U = ⊥(U⊥) and that U = (⊥U)⊥.

Let U be as described in Theorem 4.5.1. Then all the rightmost objectsof U on their respective horizontal lines form a straight half line. This isprofoundly different from Corollary 1.4.7, where given a split torsion theory

142

(X ,Y) in the finite derived category Db(mod kAn), the leftmost objects ak(on their respective lines y − x = k) which are in X form a zig zag Z. Itwould not be true if the straight half line in Theorem 4.5.1 were replaced bya half zig zag line. One example is as follows.

Example 4.5.4. Let U ′ be a subcategory of D. Suppose there is a half zig zagline such that the indecomposable objects of U ′ are precisely the objects onthe half zig zag line and to the left of it, as shown in the following diagram.Here the solid line segment on the top keeps on going to the left. The dottedhalf line is l1 : y = m (some m).

l1

U ′

In the following diagram, let U ′0 be the region outside U ′ bounded by andincluding the dotted line.

U ′ U ′0

Let U′ be the corresponding set of arcs for the subcategory U ′.

Either interpretation, as arcs or as indecomposable objects of D on thequiver, gives that U ′ is not equal to ⊥(U ′⊥). This is described as follows.

In terms of arcs, follow (with variations) Step 2 of the proof of Theorem 4.5.1(since U ′ = ⊥(U ′⊥) is equivalent to U′ satisfying conditions (?) and (? ?) byLemma 4.2.17 and Lemma 4.4.20, to have U ′ = ⊥(U ′⊥) is to have the region

143

U ′0 in U ′ as well, similar to the way the little triangular region L1 is to be inU in Theorem 4.5.1).

In terms of indecomposable objects of D on the quiver, this is simply by in-spection. The regions U ′⊥ and ⊥(U ′⊥) are as follows, i.e. the indecomposableobjects of U ′⊥ are precisely the objects on the dotted half line l2 : x = m−1and to the right of it, and the indecomposable objects of ⊥(U ′⊥) are pre-cisely the objects on the dotted half line l1 : y = m and to the left of it (c.f.Example 4.5.3).

l1 l2

⊥(U ′⊥) U ′⊥

Both interpretations give U ′ ⊂ ⊥(U ′⊥).

Remark 4.5.5. In Example 4.5.4, the set of arcs U′ does not satisfy condition(?). Therefore by Theorem 4.4.22, (U ′,U ′⊥) is not a torsion theory, eventhough U′ contains no right fountains at all. This is no surprise, as U ′ ⊂⊥(U ′⊥) contradicts Lemma 1.2.7.

Given an (indecomposable) object t of D which is neither in U ′ nor in U ′⊥,there might not even be a torsion theory triangle u′ → t→ u′′ →, where u′

is in U ′ and u′′ is in U ′⊥. Nevertheless, the following example illustrates theexistence of (some) torsion theory triangles in a t-structure explicitly.

Example 4.5.6. Consider again the t-structure (U ,V) in Theorem 4.5.1, andlet U and V be the corresponding sets of arcs for the subcategories U and Vrespectively. The lines l1 : y = m and l2 : x = m− 1, and the regions U andU⊥ are shown in the following diagram (Example 4.5.3(i)).

144

t1

l1 l2

u1

??

t2

U u2

??

q U⊥

t0

v1

u u0

??

v0y−x=2

As seen in Lemma 1.2.7, U⊥ = V and U = ⊥(U⊥). This is again no surprise(Example 4.5.3).

(i) Let t0 = (m− 2,m+ 1). Then there is the Auslander-Reiten triangleu0 → t0 → v0 →, which is also a torsion theory triangle, where u0 =(m− 2,m) is in U and v0 = (m− 1,m+ 1) is in U⊥ = V.

(ii) Let t1 = (a1, b1) be an indecomposable object of D which is neitherin U nor in U⊥. Then by Lemma 3.3.18, there is the distinguishedtriangle u1 → t1 → v1 →, which is a torsion theory triangle, whereu1 = (a1,m) is in U and v1 = (m− 1, b1) is in U⊥ = V.

(iii) Let u = (m − 4,m − 2) be in U and let t2 = (m − 4,m + 1) be anindecomposable object of D which is neither in U nor in U⊥. Thenby Lemma 3.3.18, there is the distinguished triangle u → t2 → q →,where q = (m − 3,m + 1). This is however not a torsion theorytriangle, since q is not in U⊥ = V. But similar to (ii) there is a torsiontheory triangle u2 → t2 → v2 →, where u2 = (m − 4,m) is in U andv2 = v0 = (m− 1,m+ 1) is in U⊥ = V.

(iv) It can be seen directly from the diagram that U⊥ = V is not closedunder the action of Σ (it is closed under the action of Σ−1, and it isof no surprise, see Lemma 1.2.15), and it is not true that each rightfountain of V is in fact a fountain (it is however true that each leftfountain of V is in fact a fountain, though V does not have any leftfountains).

(v) The morphisms u0 → t0 in (i), u1 → t1 in (ii) and u2 → t2 in (iii) areall U-precovers, and the morphisms t0 → v0 in (i), t1 → v1 in (ii) andt2 → v2 in (iii) are all V-preenvelopes (Example 1.2.10(ii)). CompareLemma 3.3.10 and Lemma 3.3.11.

(vi) This example echoes Remark 1.5.5 in Section 1.5. Since (U ,V) isa t-structure, the inclusion ı : U → D admits a right adjoint R :

145

D → U . Similarly, the inclusion ı : V → D admits a left adjointL : D → V (Example 1.2.25). Given t in D, there is a (unique)torsion theory triangle Rt → t → Lt → ΣRt with Rt in U and Lt inV. Accordingly, the torsion theory triangles ui → ti → vi → in (i),(ii) and (iii) serve as examples of realizing the actions of the left andright adjoints on the Auslander-Reiten quiver. For example, in (ii),Σu1 = Σ(a1,m) = (a1 − 1,m − 1) is indeed the mapping cone c1 ofthe V-preenvelope t1 → v1 (Lemma A.3.7), thus the value of Rt1 canbe retrieved (i.e. Rt1 = Σ−1c1 = Σ−1Σu1 = u1).

(vii) Simultaneously, Rt→ t in (vi) is right minimal (Lemma 1.2.11), thusin (i), (ii) and (iii), ui → ti is a U-cover.

The following example is a slight modification of Example 4.5.6.

Example 4.5.7. Let U be a subcategory of D, and U be the correspondingset of arcs for the subcategory U .

In the following diagram, suppose the indecomposable objects of U are pre-cisely the objects on the dotted half line l1 : y = m (some m) and to theleft of it, except the object . Then the indecomposable objects of U⊥ areprecisely the objects on the dotted half line l2 : x = m− 1 and to the rightof it, and the indecomposable objects of ⊥(U⊥) are precisely the objects onthe dotted half line l1 and to the left of it, thus including the object (c.f.Example 4.5.3).

l3 l1 t

l2

b0

U a u

??

U⊥

b1 v

u0 u1 u2y−x=2

Let = (m−4,m−1), a = (m−5,m−1) be objects on the line l3 : y = m−1,and u0 = (m − 6,m − 4), u1 = (m − 4,m − 2) and u2 = (m − 3,m − 1)be objects on the line y − x = 2. Even though (U ,U⊥) = 0, and as seenin Example 4.5.6(ii), that given an indecomposable object t = (a, b) of Dwhich is neither in U nor in U⊥ nor the object , there is the distinguishedtriangle u → t → v → with u = (a,m) in U and v = (m − 1, b) in U⊥, thepair (U ,U⊥) is not a torsion theory.

146

This is because there are no torsion theory triangles u → → v → withu in U and v in U⊥. Assume there is. Then since the object is in⊥(U⊥), the morphism → v is zero, and thus the distinguished triangleΣ−1v → u → → v splits, i.e. u ∼= Σ−1 v ⊕. Since U is closed underisomorphisms and direct summands, the object will have to be in U , whichis a contradiction. Therefore (U ,U⊥) is not a torsion theory.

This is again no surprise by Theorem 4.4.22, since

(i) U is not closed under extensions. This is because u1 → → u2 → is anAuslander-Reiten triangle with u1, u2 in U but not . Alternatively,let b1 = (m−6,m−3) and b0 = (m−6,m−1). Then by Lemma 3.3.18,b1 → b0 → → Σb1 is a distinguished triangle. Since both b0 and Σb1are in U but not , this is another example where U fails to be closedunder extensions.

(ii) U does not satisfy condition (?). Again consider the pair of crossingarcs u0 = (m − 6,m − 4) and a = (m − 5,m − 1), but the arc =(m− 4,m− 1) is not in U.

(iii) Simply, U ⊂ ⊥(U⊥) (Lemma 4.2.17 and Lemma 4.4.20).

The author would like to thank Rafael Bocklandt and Vanessa Miemietz forsuggesting the following example.

Let U be some subcategory of D. The following gives the example wherethe pair (U ,U⊥) is a torsion theory but not a t-structure.

Example 4.5.8. Let the indecomposable objects of U consist precisely of theobjects on the half lines y = m and x = m. Then the indecomposable objectsof U⊥ consist precisely of the objects on the dotted half lines x = m−1 andy = m− 1 (Example 4.5.3(i)).

y = m− 1 y = m x = m− 1 x = m

t0

t1

t2

Then by Theorem 4.4.22, the pair (U ,U⊥) is a torsion theory, and it isreadily seen from Theorem 4.5.1 that the pair is not a t-structure. The

147

reader can also readily establish the torsion theory triangle u0 → t0 → v0 →where u0 is in U and v0 is in U⊥ (c.f. Example 4.5.6). Similarly, by virtue ofLemma A.3.7, there is a torsion theory triangle u1 → t1 → v1 → where u1

is in U and v1 is in U⊥, and then finally by virtue of Lemma 3.3.18, there isa torsion theory triangle u2 → t2 → v2 → where u2 is in U and v2 is in U⊥.

Remark 4.5.9. By virtue of Remark 4.5.2, the subcategory U of the givenform in Theorem 4.5.1 does indeed satisfy the conditions in Theorem 4.4.22because the corresponding set of arcs U contains no right fountains at all.Since such a subcategory U is also closed under the action of Σ, the converseof Theorem 4.5.1 has to be true.

Remark 4.5.10. In Example 4.5.6, let U and V be the corresponding sets ofarcs for the subcategories U and U⊥ = V respectively. Then V = Σ ortU.This gives Σ ortU = d | Σ−1d does not cross any arcs in U = d | ddoes not cross any arcs in ΣU = ort(ΣU). Therefore ort ort(Σ ortU) =ort ort ort(ΣU) = ort(ΣU) = Σ ortU (Lemma 4.2.9). Alternatively, ort ortV =V (Lemma 4.2.17) since U⊥ = ⊥((U⊥)⊥) (Example 4.5.3).

On the other hand, V satisfies conditions (?) and (? ?), by a mirror argumentof the situation in U (Remark 4.5.2 and Remark 4.5.9). This is no surprise(Lemma 4.4.20).

By virtue of Remark 4.5.9, there are by contrast no non-trivial co-t-structures([10], [38]).

Theorem 4.5.11. Let (U ,V) be a co-t-structure in D, that is, (U ,V) is atorsion theory with Σ−1U ⊆ U . If U is non-zero, then U has to be all of D.

Proof. Let U be the corresponding set of arcs for the subcategory U . If U isnon-zero, let d = (m,n) be an indecomposable object of U .

(Step 1 ) Here it will be shown that the region R0 = (x, y)|m ≤ x, n ≤ yand y − x ≥ 2 is in U .

(i) Suppose n − m = 2. In the following diagram, let u1 = (m,n + 1).Then there is the Auslander-Reiten triangle d → u1 → Σ−1d → .Since Σ−1U ⊆ U , therefore Σ−1d is in U . Hence u1 is in U , since Uis closed under extensions. By applying a similar argument on u1 andso on, the half line x = m is in U , and so are all the half lines x = m′

with m ≤ m′, since Σ−1U ⊆ U . Therefore the region R0 is in U .

148

x = m

u1 R0

d Σ−1d

(ii) Suppose n−m > 2. Since Σ−1U ⊆ U , therefore Σ−1d = (m+1, n+1) isin U . In the following diagram, let u1 = (m,n+1) and u2 = (m+1, n).Then there is the Auslander-Reiten triangle d→ u1 ⊕ u2 → Σ−1d→.Therefore both u1 and u2 are in U , since U is closed under extensions.By applying a (similar) argument on u1 and u2 and so on (if possible),the two half lines t1 and t2 are in U . Eventually, the region R0 is inU , since Σ−1U ⊆ U .

t1

u1

d

u2 R0

t2

(Step 2 ) It follows from Step 1 that m is a right fountain of U. By Theo-rem 4.4.22, it is a fountain.

In particular, choose an indecomposable object d0 = (p,m) of U withm−p >2, as shown in the following diagram. Similar to (ii) in Step 1, the regionR1 = (x, y)|p ≤ x,m ≤ y and y − x ≥ 2 ⊃ R0 is in U .

149

y = m x = m− 2 x = m

R1

d0

(Step 3 ) Similarly, p is a right fountain of U, and so repeating the argumentof Step 2, there is a region R2 ⊃ R1 in U . Since the argument can becontinued in this way indefinitely, U has to contain all the indecomposableobjects of D, i.e. U has to be all of D.

At last we have arrived at the finale of the thesis. The following is an echoof Section 1.3 and Section 1.4 in Chapter 1.

Remark 4.5.12. Let (U ,V) be a t-structure in D.

(i) Lemma 1.3.12 and Theorem 1.4.8 together classifies t-structures in thefinite derived category Db(mod kAn). Compare Theorem 4.5.1.

(ii) Remark 1.4.12 describes the condition where the notions of split tor-sion theories and t-structures coincide (SU ∼= U). This is very far fromthe situation in D (S = Σ2).

(iii) The distinguished triangles given in Example 4.5.6(i), (ii) and (iii) areindubitably non-split, and there are no doubt indecomposable objectst in D which are neither in U nor in V. Again (U ,V) is not a splittorsion theory (Lemma 1.3.10).

Remark 4.5.13. Let (U ,V) be a co-t-structure in D.

(i) By Theorem 1.3.5, a torsion theory in D is a co-t-structure if andonly if it is a split torsion theory, since τ = SΣ−1 = Σ. However,Theorem 4.5.11 differs very much from Theorem 1.4.8, where the splittorsion theories in the finite derived category Db(mod kAn) are classi-fied.

(ii) By Lemma 1.3.10 and (i), each indecomposable d in D is either in Uor in V. This is vacuously true, since if U is non-zero, then U has tobe all of D by Theorem 4.5.11.

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(iii) By Lemma 1.3.12, every split torsion theory is a t-structure. Togetherwith (i), this means (U ,V) is a t-structure. However, this is againvacuously true, since there are no non-trivial co-t-structures by Theo-rem 4.5.11.

151

“It is a land of peace,” whispered Frodo, though he knew not from whencehis words came. “A land of quiet, and of healing, and of bright white light.There is joy there, and there is rest.”

“There is quiet and rest in the Shire,” Merry pointed out, “or at least, thereis for me.”

“Yes,” said Frodo, laughing suddenly, “and I do not want for quiet and restwhile you are with me. You are the light in my darkness, the gentle strengththat has carried me through the darkest of times. You are my family, andno more could I wish for until my work here is done.”

The white gem that hung upon Frodo’s neck flashed with a sudden light, andall four hobbits turned to see the sun rising above the mountains of Mordor,chasing away the lingering shadows with her healing rays of gold. The hob-bits stood transfigured, silver cloaks gently rippling in the soft breeze, facesshining with light, and in each eye a vision of white shores and rolling hillsof soft green grass glimmering in the golden light of day.

J. R. R. Tolkien

The Lord of the Rings

152

Appendix A

In Section A.1 and Section A.2, coherence problems in the situations of ad-joint functors and Serre functors are studied. In category theory, a multitudeof notions are portrayed in terms of commutative diagrams. As suggestedby the word coherence, certain diagrams are considered and shown to becommutative. Problems of this nature are largely contributed by the myr-iad paths connecting different Hom sets, and it is not always trivial that thedifferent paths are the same.

In Section A.3.1, some mapping cone constructions in the finite derivedcategory Db(mod kA7) are illustrated. It is an attempt to provide an alter-native to the mapping cone construction in Section 3.2.2, in the manner ofSection 3.3.2.

In Section A.3.1 and Section A.3.2, mapping cones of (compositions of)downward morphisms in Db(mod kA7) and the cluster category of Dynkintype A∞ are also portrayed. They serve to enrich our repertoire of map-ping cone constructions, albeit the different suggestiveness bequeathed uponthem.

A.1 Coherence problems for adjoint functors

In this section, let A and B be arbitrary categories, and let the pair offunctors L : A → B and R : B → A be adjoints, i.e. there is an isomorphism

τ : (LA,B)∼=→ (A,RB) for all A in A and B in B, such that τ is natural in

A and B. This is given in Example 0.1.17(iv).

Let A be in A and B be in B. Then the map ηA = τ(idLA) : A → RLA isthe unit of the adjunction, and the map εB = τ−1(idRB) : LRB → B is thecounit of the adjunction.

Example A.1.1. Let A be in A and B be in B. Suppose ηA is invertible, andconsider ηRB = τ(idLRB) : RB → RLRB. Then η−1

RB = R(εB) : RLRB →

153

RB.

Proof. Since τ is natural in B, there is the following commutative diagram.

(LRB,LRB)τRB,LRB //

(LRB,εB)

(RB,RLRB)

(RB,RεB)

(LRB,B)

τRB,B // (RB,RB)

Therefore (RB,RεB)τRB,LRB(idLRB) = τRB,B(LRB, εB)(idLRB).

Thus (RB,RεB)ηRB = τRB,BεB,

and finally, RεBηRB = idRB.

Since ηRB is invertible, η−1RB = R(εB).

Example A.1.2. Let A be in A and B be in B. Suppose A and B aretriangulated with translation functors ΣA and ΣB respectively. If (L, σL) :

A → B is a triangulated functor where σL : LΣA'→ ΣBL, then by [35,

Lemma 5.3.6], (R, σR) is also a triangulated functor where σR : RΣB'→

ΣAR.

Consider εΣBB : LRΣBB → ΣBB and ΣB(εB) : ΣBLRB → ΣBB. Since

σRB : RΣBB∼=→ ΣARB, therefore LσRB : LRΣBB

∼=→ LΣARB. Also σLRB :

LΣARB∼=→ ΣBLRB. Is the following diagram commutative? The author at

the time of writing is not able to give an answer.

LRΣBBεΣBB //

L(σRB)

ΣBB

L(ΣAR)B

(LΣA)RB

σLRB

ΣBLRBΣB(εB)

// ΣBB

154

A.2 Coherence problems for Serre functors

In this section, let A be a k-linear and Hom finite additive category with aSerre functor S. This is given in Definition 0.3.8. Nevertheless, it is statedas follows.

Notation A.2.1. A right Serre functor is an additive functor S : A → A,together with isomorphisms

ϕA,B : (A,B)∼=→ (B,SA)∗

for any A,B ∈ A, which are natural in A and in B, and where (−)∗ =Homk(−, k). If S is an autoequivalence, then it is a Serre functor.

Let ϕA,A(idA) be denoted by ϕA. The notations ϕA and ϕA,B, where A,B ∈A, will be used for the rest of the section.

Let V and W be k-vector spaces and consider f : V → W . Then for θ inW ∗ and v in V , there is a natural way to define f∗ : W ∗ → V ∗ described asfollows.

f∗(θ)(v) = θ(f(v)) (A.2.1)

Given the k-vector spaces U , V and W , consider f : U → V and g : V →W .Then by equation (A.2.1), we have

(gf)∗ = f∗g∗. (A.2.2)

Notation A.2.2. Let U be a finite-dimensional k-vector space. Then there

is a natural way to define the double dual isomorphism ηU : U∼=→ U∗∗, i.e.

for u in U , f in U∗,ηU (u)(f) = f(u). (A.2.3)

The notation ηU , simply written η, where U is a finite-dimensional k-vectorspace, will be used for the rest of the section.

The isomorphism η is natural in the following sense. Suppose U and Vare finite-dimensional k-vector spaces. Then given f : U → V , there is thefollowing commutative diagram.

U

f

η // U∗∗

f∗∗

Vη // V ∗∗

155

This is because given ρ in V ∗, u in U , we have

f∗∗(η(u))(ρ) = η(u)(f∗(ρ)) (by equation (A.2.1))

= f∗(ρ)(u) (by equation (A.2.3))

= ρ(f(u)). (by equation (A.2.1))

On the other hand, since

η(f(u))(ρ) = ρ(f(u)), (by equation (A.2.3))

thereforef∗∗(η(u)) = η(f(u)). (A.2.4)

This means the diagram is commutative.

Example A.2.3. Let A, B be in A. Consider ϕB,SA : (B,SA)∼=→ (SA, SB)∗,

η : (SA, SB)∼=→ (SA, SB)∗∗ and f : A → B, g : B → SA in A. Then we

have

(ϕ∗B,SAη(Sf))(g) = (η(Sf))(ϕB,SA(g)) (by equation (A.2.1))

= (ϕB,SA(g))(Sf). (by equation (A.2.3))

Therefore(ϕ∗B,SAη(Sf))(g) = (ϕB,SA(g))(Sf). (A.2.5)

Example A.2.4. Let A1, A2, B be in A. Consider ϕAi,B : (Ai, B)∼=→

(B,SAi)∗, i = 1, 2. Consider g : A1 → A2, θ : A2 → B and f : B → SA1

in A. Since ϕAi,B is natural in the first variable, there is the followingcommutative diagram.

(A2, B)

(g,B)

ϕA2,B// (B,SA2)∗

(B,Sg)∗

(A1, B) ϕA1,B

// (B,SA1)∗

This is to say

156

ϕA1,B (g,B) = (B,Sg)∗ ϕA2,B, (A.2.6)

which gives

(ϕA1,B(θg))(f) = ϕA2,B(θ)((Sg)f).

Suppose A2 = B and θ = idB, then

(ϕA1,B(g))(f) = ϕB,B(idB)((Sg)f) = ϕB((Sg)f). (A.2.7)

Finally, A.2.2 and A.2.6 together give

(g,B)∗ ϕ∗A1,B = ϕ∗A2,B (B,Sg)∗∗. (A.2.6*)

Example A.2.5. Let A, B1, B2 be in A. Consider ϕA,Bi : (A,Bi)∼=→

(Bi, SA)∗, i = 1, 2. Consider g : B1 → B2, θ : A → B1 and f : B2 → SAin A. Since ϕA,Bi is natural in the second variable, there is the followingcommutative diagram.

(A,B1)

(A,g)

ϕA,B1// (B1, SA)∗

(g,SA)∗

(A,B2) ϕA,B2

// (B2, SA)∗

This is to say

ϕA,B2 (A, g) = (g, SA)∗ ϕA,B1 , (A.2.8)

which gives

(ϕA,B2(gθ))(f) = ϕA,B1(θ)(fg).

Suppose A = B1 and θ = idB1 , then

ϕB1,B2(g)(f) = ϕB1,B1(idB1)(fg) = ϕB1(fg). (A.2.9)

Finally, A.2.2 and A.2.8 together give

(A, g)∗ ϕ∗A,B2= ϕ∗A,B1

(g, SA)∗∗. (A.2.8*)

157

Lemma A.2.6. Let A, B be in A, and consider f : A→ B and g : B → SAin A. Then ϕB,SA(g)(Sf) = ϕA,B(f)(g).

Proof.

ϕB,SA(g)(Sf) = ϕB((Sf)g) (by equation (A.2.9))

= ϕA,B(f)(g). (by equation (A.2.7))

The first equality corresponds to the lower commutative square, while thesecond equality corresponds to the upper commutative square in the follow-ing diagram.

(A,B)ϕA,B // (B,SA)∗

(B,B)

(f,B)

OO

(B,g)

ϕB,B // (B,SB)∗

(B,Sf)∗

OO

(g,SB)∗

(B,SA) ϕB,SA

// (SA, SB)∗

Lemma A.2.7. Let A, B be in A, and consider g : A→ B and θ : B → SAin A. Then (ϕSA,SB(Sg))(Sθ) = ϕA,B(g)(θ).

Proof.

(ϕSA,SB(Sg))(Sθ) = ϕSA(SθSg) (by equation (A.2.9))

= ϕB,SA(θ)(Sg) (by equation (A.2.7))

= ϕB((Sg)θ) (by equation (A.2.9))

= ϕA,B(g)(θ). (by equation (A.2.7))

(A,B)

S(−,−)

ϕA,B // (B,SA)∗

(SA, SB)ϕSA,SB// (SB, S2A)∗

S(−,−)∗

OO

158

Lemma A.2.8. Let A, B be in A. Consider η : (SA, SB)∼=→ (SA, SB)∗∗

and then f : A → B and g : B → SA in A. Then (ϕ∗B,SAη(Sf))(g) =ϕA,B(f)(g).

Proof.

(ϕ∗B,SAη(Sf))(g) = ϕB,SA(g)(Sf) (by equation (A.2.5))

= ϕA,B(f)(g). (by Lemma A.2.6)

(A,B)

S(−,−)

ϕA,B // (B,SA)∗

(ϕ∗B,SA)−1

(SA, SB)

η // (SA, SB)∗∗

Therefore ϕ∗B,SAη(Sf) = ϕA,B(f).

Lemma A.2.9. Let A, B be in A. Consider η : (SA, SB)∼=→ (SA, SB)∗∗

and then g : A → B and f : B → SA in A. Then (ϕ∗B,SAη(Sg))(f) =ϕSA,SB(Sg)(Sf).

Proof.

(ϕ∗B,SAη(Sg))(f) = ϕA,B(g)(f) (by Lemma A.2.8)

= ϕSA,SB(Sg)(Sf). (by Lemma A.2.7)

(SB, S2A)∗

S(−,−)∗

(ϕSA,SB)−1

// (SA, SB)

η

(B,SA)∗ (SA, SB)∗∗

ϕ∗B,SA

oo

Example A.2.10. Consider again the diagram in Lemma A.2.7.

(A,B)

S(−,−)

ϕA,B // (B,SA)∗

(SA, SB) ϕSA,SB

// (SB, S2A)∗

S(−,−)∗

OO

159

Consider η : (SA, SB)∼=→ (SA, SB)∗∗. The diagram can be expressed as

follows.

(A,B)

S(−,−)

ϕA,B // (B,SA)∗

(SA, SB)η //

ϕSA,SB

(SA, SB)∗∗

ϕ∗B,SA

OO

ϕ∗B,SA

(SB, S2A)∗

S(−,−)∗// (B,SA)∗

The upper square corresponds to the diagram in Lemma A.2.8, while thelower square corresponds to the diagram in Lemma A.2.9.

Lemma A.2.11. Let A, B, Y be in A, and consider y : SY → B in A.Then the following diagram is commutative.

(B,SA)η //

(y,SA)

(B,SA)∗∗ϕ∗A,B // (A,B)∗

(A,y)∗

(SY, SA)

(S(−,−))−1

(Y,A) ϕY,A

// (A,SY )∗

Proof. The diagram can be decomposed as follows.

(B,SA)η //

(y,SA)

(B,SA)∗∗ϕ∗A,B //

(y,SA)∗∗

(A,B)∗

(A,y)∗

(SY, SA)

(S(−,−))−1

η // (SY, SA)∗∗

ϕ∗A,SY

''(Y,A) ϕY,A

// (A,SY )∗

The upper left square commutes by equation (A.2.4). The square on theright commutes by equation (A.2.8*). The square on the bottom commutesby Lemma A.2.8.

160

A.3 Mapping cone construction

The author would like to thank Yann Palu for suggesting the constructions ofmapping cones of (compositions of) downward morphisms in Db(mod kA7)and in the cluster category of Dynkin type A∞.

A.3.1 The finite derived category Db(mod kA7)

This section considers only the finite derived category Db(mod kA7), never-theless the lemmas can be readily generalized in the finite derived categoryDb(mod kAn). As usual, let Σ be the translation functor of Db(mod kA7).

Lemma A.3.1 and Lemma A.3.2 in this section have been mentioned inRemark 3.3.21.

Consider the Auslander-Reiten quiver of the finite derived categoryDb(mod kA7)in the following diagram.

•

•

•

•

•

•

•

•

•

??

•

??

•

??

•

??

•

??

•

??

•

??

. . . •

??

•

??

•

??

•

??

•

??

•

??

•??

• . . .

•

??

•b2

??

•

??

•

??

•

??

•

??

•

??

•??

•a2

f2 ??

g2•b1

??

•??

•??

•??

•??

•

•??

•a1

f1 ??

g1•b0

??

•??

•??

•??

•??

•??

•??

•a0

f0 ??

•c??

•??

•??

•??

•

The coordinates of some of the objects are as follows: a0 = (i, j), b0 =(i, j + 1), a1 = (i− 1, j), b1 = (i− 1, j + 1), a2 = (i− 2, j), b2 = (i− 2, j + 1)and c = (i + 1, j + 1). As usual, the coordinates of objects on the bottomline satisfy the equation y − x = 2.

Lemma A.3.1. (c.f. Lemma 3.3.16) The mapping cones of the maps fn :(i− n, j)→ (i− n, j + 1) are all isomorphic to c = (i+ 1, j + 1).

Proof. This is similar to Lemma 3.3.16. The two categories, the finite de-rived category Db(mod kAn) and the cluster category of Dynkin type A∞,are distinct, though their Auslander-Reiten quivers have some similarity.The action of the translation functor Σ is not relevant in this situation.Nevertheless, a proof is given below. Since a0 → b0 → c→ is an Auslander-Reiten triangle, the statement is true for n = 0. Suppose the statement is

161

true for n = p, p ≥ 0, i.e. the mapping cone of the map fp : ap → bp isc = (i+ 1, j + 1). To see that the statement is true for n = p+ 1, considerthe following commutative diagram,

ap+1µp+1 // ap ⊕ bp+1

ap+1

fp+1

// bp+1,

where µp+1 =

(gp+1

fp+1



0 //

ap

ı

ap

fp

// 0

ap+1

µp+1 // ap ⊕ bp+1//

bp

// Σap+1

ap+1fp+1

//

bp+1

// c

// Σap+1

0 // Σap Σap // 0,

where the map ı is the canonical injection. The distinguished triangle onthe second row is an Auslander-Reiten triangle. Hence the mapping cone ofthe map fp+1 : ap+1 → bp+1 is the same as the mapping cone of the mapfp : ap → bp, which is c = (i+ 1, j + 1) by the induction hypothesis.

The following is similar to Lemma 3.3.18, and the proof is left to the reader.The action of Σ is implicitly alluded and is not entirely irrelevant in thissituation (c.f. Lemma A.3.1). This will be illustrated in Example A.3.3.

Lemma A.3.2. (c.f. Lemma 3.3.18) Let 3 ≤ j − i ≤ 8 and consider thefollowing sketch where the Auslander-Reiten quiver of Db(mod kA7) lies inthe region bounded by y − x = 2 and y − x = 8.

162

y−x=8

•(i,j)

!!

•(i,j−r−1)

==

•cr

•c0 y−x=2

For 0 ≤ r ≤ j − i − 2, let ar = (i, j − r). Let fr be the morphism fr :(i, j−r−1)→ (i, j−r), where 0 ≤ r ≤ j− i−3. Then for 0 ≤ r ≤ j− i−3,the composition f0 . . . fr−1fr : (i, j − r − 1)→ (i, j) has mapping cone cr =(j − r − 2, j).

Example A.3.3. Consider the following sketch.

•(i,i+8)

y−x=8

•(i,i+7−r)

??

•cr

•c0 y−x=2

For −1 ≤ n ≤ 5, let an = (i, i + 7 − n). Let fn be the morphism fn :(i, i + 7 − n) → (i, i + 8 − n), where 0 ≤ n ≤ 5. Then for 0 ≤ r ≤ 5,the composition f0 . . . fr−1fr : (i, i + 7 − r) → (i, i + 8) has mapping conecr = (i+ 6− r, i+ 8).

Proof. Consider the object c0 = (i + 6, i + 8) on the bottom line y − x =2. It lies in the same descending line as (i, i + 8), where the compositionf0 . . . fr−1fr maps to. The mapping cone cr = (i + 6 − r, i + 8) lies in thesame descending line as c0, but it is r steps up the line. This is how weunderstand the location of the mapping cone cr.

163

Consider the Auslander-Reiten quiver of the finite derived categoryDb(mod kA7).

•

••a−1

•

•

•

•

•

•

??

•a0

f0 ??

•

??

•

??

•

??

•

??

•

??

. . . •

??

•a1

f1 ??

•

??

•

??

•

??

•

??

•??

• . . .

•a2

f2 ??

•

??

•

??

•

??

•Σa2

??

•

??

•

??

•??

•??

•??

•??

•??

•Σa1

??

•??

•

•??

•??

•??

•??

•c1

??

•Σa0

??

•??

•??

•??

•??

•??

•d0

??

•c0??

•Σa−1

??

•

The statement is true for r = 0 by Lemma A.3.2. Suppose the statement istrue for r = p, p ≥ 0, and then the statement is also required to be true forr = p+ 1.


ap+1fp+1 // ap

g

ap+1

gfp+1

// a−1,

where g is the morphism f0 . . . fp : (i, i+ 7− p)→ (i, i+ 8).


ap+1fp+1 // ap

g

ρ // dp

// Σap+1

ap+1gfp+1 //

a−1//

∗

// Σap+1

0 //

cp

%

cp

(Σρ)%

// 0

Σap+1

// ΣapΣρ // Σdp // Σ2ap+1.

By the induction hypothesis, the mapping cone of the morphism g is cp =(i+6−p, i+8), which gives the distinguished triangle in the second column.By Lemma A.3.1, the mapping cone of the morphism fp+1 is the objectdp = (i+ 5− p, i+ 7− p). This gives the distinguished triangle on the firstrow.

164

The map ρ : ap → dp is non-zero, as otherwise ap+1∼= Σ−1dp ⊕ ap by

Lemma 0.2.2(v), which is not possible since ap+1 is indecomposable. There-fore the map Σρ : Σap → Σdp is non-zero as well. Similarly, the map% : cp → Σap is non-zero. For −1 ≤ p ≤ 4, dp lies on the line y = i + 7 − pand hence Σdp lies on the line x = i + 6 − p. This gives Σdp in R(cp).Therefore the composition (Σρ)% : cp → Σdp is non-zero by Lemma 3.2.3,and the distinguished triangle dp → ∗ → cp → Σdp is non-split.

Let an object e have coordinates (i + 5 − p, i + 8). By Lemma A.3.2, themapping cone of dp → e is cp. Since (cp,Σdp) is one-dimensional, the object∗ is indeed equal to e. Therefore the mapping cone of the morphism gfp+1

is e = (i+ 5− p, i+ 8) = (i+ 6− (p+ 1), i+ 8) = cp+1 as desired.

The following two lemmas consider mapping cones of (compositions of)downward morphisms, and they are delivered entirely out of leisure. Theirproofs are the mirror versions of the proofs of Lemma A.3.1 and Lemma A.3.2respectively, since the action of the translation functor Σ corresponds to theglide reflection. Nevertheless, they are given below, so that the reader canbe guided formally and rigorously with good intuition.

They also engender yet another significance in the construction of distin-guished triangles in the quotient categories. This is similar to the forthcom-ing section, and the reader can refer to the description given there.

Consider again the Auslander-Reiten quiver in the following diagram.

•

•

••a−1

f0•b−1

•

•

•

•

??

•

??

•a0f1

g−1 ??

•b0

??

•

??

•

??

•

??

. . . •

??

•

??

•a1f2

g0 ??

•b1

??

•

??

•

??

•??

• . . .

•

??

•a2f3

g1 ??

•b2

??

•

??

•

??

•

??

•

??

•??

•??

•b3

??

•??

•??

•??

•??

•

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•

For −1 ≤ n ≤ 5, let an = (i, i+ 7− n), bn = (i+ 1, i+ 8− n).

Lemma A.3.4. (c.f. Lemma A.3.1)

The mapping cones of the maps fn : (i, i+ 8− n)→ (i+ 1, i+ 8− n) are allisomorphic to b−1 = (i+ 1, i+ 9).

165

Proof. Since a−1 → b0 → b−1 → is an Auslander-Reiten triangle, the state-ment is true for n = 0. Suppose the statement is true for n = p, p ≥ 0, i.e.the mapping cone of the map fp : ap−1 → bp is b−1 = (i + 1, i + 9). To seethat the statement is true for n = p+ 1, consider the following commutativediagram,

apµp−1 // ap−1 ⊕ bp+1

ap

fp+1

// bp+1,

where µp−1 =

(gp−1

fp+1



0 //

ap−1

ı

ap−1

fp

// 0

ap

µp−1 // ap−1 ⊕ bp+1//

bp

// Σap

apfp+1

//

bp+1

// c

// Σap

0 // Σap−1 Σap−1

// 0,

where the map ı is the canonical injection. The distinguished triangle onthe second row is an Auslander-Reiten triangle. Hence the mapping coneof the map fp+1 : ap → bp+1 is the same as the mapping cone of the mapfp : ap−1 → bp, which is b−1 = (i+ 1, i+ 9) by the induction hypothesis.

Lemma A.3.5. (c.f. Lemma A.3.2, Example A.3.3)

Consider the following sketch.

166

•a−1

•b−1y−x=8

•cr

•ar

y−x=2

For −1 ≤ n ≤ 5, let an = (j − 7 + n, j) and bn = (j − 6 + n, j + 1). Let fnbe the morphism fn : (j − 8 + n, j)→ (j − 7 + n, j), where 0 ≤ n ≤ 5. Thenfor 0 ≤ r ≤ 5, the composition fr . . . f1f0 : (j − 8, j) → (j − 7 + r, j) hasmapping cone cr = br−1 = (j − 7 + r, j + 1).

Proof. Consider the object b−1 = (j − 7, j + 1) on the top line y − x = 8.The mapping cone cr = (j − 7 + r, j + 1) lies in the same descending line asb−1, but it is r steps down the line. The object b−1 lies immediately to theright of a−1, where a−1 is whence the composition fr . . . f1f0 maps from.This is how we understand the location of the mapping cone cr.

Consider again the Auslander-Reiten quiver in the following diagram.

••a−1

f0•b−1

•

•

•

•

•

•

??

•a0f1

??

•

??

•

??

•

??

•

??

•

??

. . . •

??

•

??

•a1f2

??

•

??

•

??

•

??

•??

• . . .

•

??

•

??

•a2

??

•

??

•

??

•

??

•

??

•??

•??

•??

•??

•??

•??

•??

•

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•??

•


167


a−1g // ap

fp+1

a−1

fp+1g// ap+1,

where g is the morphism fp . . . f0 : (j − 8, j)→ (j − 7 + p, j).


a−1g // ap

fp+1

ρ // cp

// Σa−1

a−1fp+1g //

ap+1//

∗

// Σa−1

0 //

dp

%

dp

(Σρ)%

// 0

Σa−1

// ΣapΣρ // Σcp // Σ2a−1.

The distinguished triangle on the first row is given by the induction step.The mapping cone of the morphism fp+1 is dp = (j − 6 + p, j + 2 + p) byLemma A.3.4, which gives the distinguished triangle in the second column.

The map ρ : ap → cp is non-zero, as otherwise a−1∼= Σ−1cp ⊕ ap by

Lemma 0.2.2(v), which is not possible since a−1 is indecomposable. There-fore the map Σρ : Σap → Σcp is non-zero as well. Similarly, the map% : dp → Σap is non-zero. Since Σcp = (j, j+2+p) lies on the line y = j+2+p,this gives Σcp in R(dp). Therefore the composition (Σρ)% : dp → Σcp is non-zero by Lemma 3.2.3, and the distinguished triangle cp → ∗ → dp → Σcp isnon-split.

Let an object e have coordinates (j − 6 + p, j + 1). By Lemma A.3.4, themapping cone of cp → e is dp. Since (dp,Σcp) is one-dimensional, the object∗ is indeed equal to e. Therefore the mapping cone of the morphism fp+1gis e = (j − 6 + p, j + 1) = (j − 7 + (p+ 1), j + 1) = cp+1 as desired.

A.3.2 The cluster category of Dynkin type A∞

This section considers mapping cones of (compositions of) downward (for-ward) morphisms in the cluster category D of Dynkin type A∞. As usual,let Σ be the translation functor of D.

168

Given a subcategory X of D which is both precovering and preenveloping,the quotient category DX is pretriangulated. DX is triangulated if and onlyif τX = X , where τ is the Auslander-Reiten translation. This is describedin Section 3.1.

The distinguished triangles in DX are described in [23, Setup 1.1]. Theyare obtained either by apprehending X as preenveloping or as precovering.If X is perceived as preenveloping, then Lemma 3.3.16 and Lemma 3.3.18are needed (Example 3.3.19). In a different manner, if X is perceived asprecovering, then the following lemmas are needed (Example A.3.8).

Consider the Auslander-Reiten quiver of D in the following diagram.

...

•

•

•

•

•

•

??

•

??

•

??

•

??

•

??

•

•

??

•

??

•a3g2

??

•

??

•

??

. . . •

??

•

??

•a2g1

f2 ??

•b2

??

•

??

• . . .

•

??

•a1g0

f1 ??

•b1

??

•

??

•

??

•c??

•a0

f0 ??

•b0??

•

??

•

??

•

Let us write down the coordinates of some of the objects shown: a0 =(i, j), b0 = (i+ 1, j+ 1), a1 = (i, j+ 1), b1 = (i+ 1, j+ 2), a2 = (i, j+ 2), b2 =(i+ 1, j + 3) and c = (i− 1, j − 1). As usual, the coordinates of objects onthe bottom line satisfy the equation y − x = 2.

Lemma A.3.6. (c.f. Lemma 3.3.16, Lemma A.3.4) The mapping conesof the maps gn : (i, j + n + 1) → (i + 1, j + n + 1) are all isomorphic toc = Σa0 = (i− 1, j − 1).

Proof. Since a0 → a1 → b0 → Σa0 is an Auslander-Reiten triangle, thestatement is true for n = 0. Suppose the statement is true for n = p, p ≥ 0,i.e. the mapping cone of the map gp : ap+1 → bp is c = (i− 1, j − 1). To seethat the statement is true for n = p+ 1, consider the following commutativediagram,

ap+1µp+1 // ap+2 ⊕ bp

ap+1 gp

// bp,

169

where µp+1 =

(fp+1

gp



0 //

ap+2

ı

ap+2

gp+1

// 0

ap+1

µp+1 // ap+2 ⊕ bp //

bp+1

// Σap+1

ap+1 gp//

bp

// c

// Σap+1

0 // Σap+2 Σap+2

// 0,

where the map ı is the canonical injection. The distinguished triangle onthe second row is an Auslander-Reiten triangle. Hence the mapping cone ofthe map gp+1 : ap+2 → bp+1 is the same as the mapping cone of the mapgp : ap+1 → bp, which is c = (i− 1, j − 1) by the induction hypothesis.

Lemma A.3.7. (c.f. Lemma 3.3.18, Lemma A.3.5) Consider the followingsketch.

•Σ(i,j) •(i,j)

•cr •(i+r+1,j)

•c0y−x=2

For 0 ≤ n, let an = (i+ n, j). Let gn be the morphism gn : (i+ n, j)→ (i+n+1, j). Then for 0 ≤ r, the composition grgr−1 . . . g0 : (i, j)→ (i+r+1, j)has mapping cone cr = (i− 1, i+ 1 + r).

Proof. Consider the object c0 = (i− 1, i+ 1) on the bottom line y − x = 2.It lies in the same ascending line as Σ(i, j). The mapping cone cr = (i −1, i+1 + r) lies in the same ascending line as c0, but it is r steps up the line.This is how we understand the location of the mapping cone cr.

170

Consider again the Auslander-Reiten quiver of D.

...

•

•

•

•

•

•

•

•

??

•

??

•Σa0

??

•a0g0

??

•

??

•

??

. . . •

??

•

??

•

??

•Σa1

??

•a1g1

??

•

??

• . . .

•

??

•

??

•

??

•

??

•a2g2

??

•

??

•

??

•??

•??

•??

•??

•a3

??

•

•c1

??

•??

•??

•??

•??

•??

•c0

??

•??

•??

•??

•??

•??

•



a0g // ap+1

gp+1

a0 gp+1g

// ap+2,

where g is the morphism gpgp−1 . . . g0 : (i, j)→ (i+ p+ 1, j).


a0g // ap+1

gp+1

ρ // dp

// Σa0

a0 gp+1g//

ap+2//

∗

// Σa0

0 //

ep

%

ep

(Σρ)%

// 0

Σa0

// Σap+1Σρ // Σdp // Σ2a0.

Again consider the object c0 = (i− 1, i+ 1) on the bottom line y−x = 2. Itlies in the same ascending line as Σa0, where a0 is the place from where themorphism g maps. By the induction hypothesis, the mapping cone of themorphism g lies in the same ascending line as c0, but it is p steps up the line,

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i.e. dp = cp = (i− 1, i+ 1 + p). This gives the distinguished triangle on thefirst row. The mapping cone of the morphism gp+1 is ep = (i+p, i+p+2) byLemma A.3.6, which gives the distinguished triangle in the second column.

The map ρ : ap+1 → dp is non-zero, as otherwise a0∼= Σ−1dp ⊕ ap+1 by

Lemma 0.2.2(v), which is not possible since a0 is indecomposable. Thereforethe map Σρ : Σap+1 → Σdp is non-zero as well. Similarly, the map % : ep →Σap+1 is non-zero. Therefore the composition (Σρ)% : ep → Σdp is non-zeroby Lemma 3.3.8, and the distinguished triangle dp → ∗ → ep → Σdp isnon-split.

Let an object s have coordinates (i − 1, i + 2 + p). By Lemma 3.3.16, themapping cone of dp → s is ep. Since (ep,Σdp) is one-dimensional, the object∗ is indeed equal to s. Therefore the mapping cone of the morphism gp+1gis ∗ = s = (i− 1, i+ 2 + p) = (i− 1, i+ 1 + (p+ 1)) = cp+1 as desired.

The section finishes with the following example which illustrates the obtain-ing of the distinguished triangles in the quotient category DX in virtue ofX -precovers. The reader can refer to [23, Setup 1.1] for more details.

Example A.3.8. (c.f. Example 3.3.19) Consider again the Auslander-Reitenquiver and the subcategory X in Example 3.3.19. As usual, let σ be thetranslation functor of DX .

...

•

•

•

•

•

•

•

•

??

•

??

•b−1g−1

??

•a−1

??

•

??

•

??

. . . •

??

•

??

•b0

??

•a0g0

??

•

??

•

??

• . . .

•Σb1

??

•b1

??

•a1

??

•d3g1

??

•c3

??

•

??

•

??

•b2??

•a2

??

•

??

•d2

??

g2•c2

??

•

•b3

??

•a3

??

•

??

•

??

•d1

??

g3•c1

??

•b4

??

•a4

??

•

??

•

??

•

??

•d0

??

•c0

For example, the morphism g = g2g1g0 : a0 → d1 is an X -epimorphism in D,i.e. a morphism such that each morphism x→ d1 with x in X factors throughg (c.f. Lemma 3.3.11). Extend g : a0 → d1 to the distinguished triangle

a0g→ d1 → b2 → Σa0 in D by Lemma A.3.7, i.e. b2 is the mapping cone

of the morphism g. On the other hand, the mapping cone of the morphismgg−1 is Σb1 by Lemma A.3.7, where gg−1 : b−1 → d1 is an X -precover of

172

d1 by Lemma 3.3.11. Then the diagram Σ−1Σb1 → Σ−1b2 → a0 → d1,which is the diagram b1 → a2 → a0 → d1, considered in DX , is definedto be a distinguished triangle in DX , so that σ−1(d1) = b1. Compare inExample 3.3.19(iii) the distinguished triangle a2 → a0 → d1 → c2 in DX .

173

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